Limits and Continuity

Introduction

In calculus, one of the most important ideas is the limit. A limit describes what value a function is approaching as the input gets closer and closer to a certain number.

A limit does not always tell us the actual value of the function at that number. Instead, it tells us about the function’s behavior nearby. This makes limits very useful when studying graphs, holes, asymptotes, boundaries, and continuity.

For example, if we want to know what happens to a function as \(x\) gets close to \(2\), we write

$$\lim_{x \to 2} f(x)$$

This is read as: “the limit of \(f(x)\) as \(x\) approaches 2.”

1. What a Limit Means Intuitively

Suppose a function has values that get closer and closer to \(5\) when \(x\) gets closer and closer to \(2\). Then we say

$$\lim_{x \to 2} f(x)=5$$

Even if \(f(2)\) is not equal to \(5\), the limit can still be \(5\). The key idea is what happens near \(x=2\), not necessarily at \(x=2\).

This is why limits help us study functions with:

• holes in the graph,
• jumps,
• vertical asymptotes,
• end behavior near boundaries.

2. Looking at Limits with a Table

Consider the function

$$f(x)=\frac{x^2-4}{x-2}$$

If we substitute \(x=2\), we get

$$\frac{2^2-4}{2-2}=\frac{0}{0}$$

This is undefined, so the function is not defined at \(x=2\). But we can still study the limit by checking values near \(2\).

Since \(x^2-4=(x-2)(x+2)\), for \(x\neq 2\) we have

$$f(x)=x+2$$

Now make a table of values near \(2\):

From the left of \(2\):

• if \(x=1.9\), then \(f(x)=3.9\)
• if \(x=1.99\), then \(f(x)=3.99\)
• if \(x=1.999\), then \(f(x)=3.999\)

From the right of \(2\):

• if \(x=2.1\), then \(f(x)=4.1\)
• if \(x=2.01\), then \(f(x)=4.01\)
• if \(x=2.001\), then \(f(x)=4.001\)

These values get closer to \(4\), so

$$\lim_{x \to 2} \frac{x^2-4}{x-2}=4$$

Although the function is undefined at \(x=2\), the limit still exists.

3. Looking at Limits on a Graph

On a graph, a limit asks: What y-value is the graph approaching as x gets close to a certain number?

If the graph approaches the same y-value from both sides, the limit exists.

If the left side and right side approach different values, the limit does not exist.

If the graph rises or falls without bound near a value of \(x\), the function may have a vertical asymptote. In that case, the limit may be infinity or negative infinity.

4. Left-Hand and Right-Hand Limits

Sometimes we must check each side separately.

The left-hand limit is written as

$$\lim_{x \to a^-} f(x)$$

This means \(x\) approaches \(a\) from values less than \(a\).

The right-hand limit is written as

$$\lim_{x \to a^+} f(x)$$

This means \(x\) approaches \(a\) from values greater than \(a\).

A two-sided limit exists only if both one-sided limits exist and are equal:

$$\lim_{x \to a} f(x)=L \quad \text{if and only if} \quad \lim_{x \to a^-} f(x)=L \text{ and } \lim_{x \to a^+} f(x)=L$$

5. Limits and Actual Function Values Are Different Ideas

It is important to separate these three questions:

1. What is \(f(a)\)?
2. What is \(\lim_{x \to a} f(x)\)?
3. Are they equal?

Sometimes all three work nicely together. Sometimes they do not.

For a function to be continuous at \(x=a\), three things must happen:

1. \(f(a)\) exists,
2. \(\lim_{x \to a} f(x)\) exists,
3. \(\lim_{x \to a} f(x)=f(a)\).

This shows why limits are the foundation of continuity.

6. Formal Definition of a Limit

The formal idea says that \(\lim_{x \to a} f(x)=L\) means we can make \(f(x)\) as close as we want to \(L\) by taking \(x\) close enough to \(a\), but not necessarily equal to \(a\).

In symbols, this is written as:

For every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if

$$0<|x-a|<\delta$$

then

$$|f(x)-L|<\varepsilon$$

You do not need to be afraid of this notation. Its meaning is simple:

• \(\varepsilon\) tells us how close we want the output to be to \(L\),
• \(\delta\) tells us how close the input must be to \(a\).

So the formal definition turns the intuitive idea of “approaching” into an exact mathematical statement.

7. Common Ways to Evaluate Limits

In 12th Grade, limits are often found using these methods:

• Direct substitution: if the function is continuous at that point, just substitute the value.
• Algebraic simplification: factor, cancel, or simplify first.
• Numerical tables: examine values close to the point.
• Graphs: inspect what the function approaches.
• One-sided limits: check left and right behavior separately.

Worked Example 1: A Basic Continuous Function

Evaluate

$$\lim_{x \to 3}(2x+1)$$

Step 1: This is a linear function, which is continuous everywhere.

Step 2: Use direct substitution.

$$\lim_{x \to 3}(2x+1)=2(3)+1=7$$

Answer: \(7\)

Because the function is continuous, the limit equals the function value.

Worked Example 2: A Limit with a Hole

Evaluate

$$\lim_{x \to 2}\frac{x^2-4}{x-2}$$

Step 1: Direct substitution gives

$$\frac{2^2-4}{2-2}=\frac{0}{0}$$

This is indeterminate, so we simplify.

Step 2: Factor the numerator.

$$x^2-4=(x-2)(x+2)$$

So for \(x\neq 2\),

$$\frac{x^2-4}{x-2}=x+2$$

Step 3: Now take the limit.

$$\lim_{x \to 2}(x+2)=4$$

Answer: \(4\)

The graph looks like the line \(y=x+2\) with a hole at \((2,4)\). The function value at \(2\) is missing, but the limit is still \(4\).

Worked Example 3: One-Sided Limits and a Jump

Consider the piecewise function

$$f(x)=\begin{cases}x+1,&x<2\\5,&x\ge 2\end{cases}$$

Find \(\lim_{x \to 2}f(x)\).

Step 1: Find the left-hand limit.

For \(x<2\), \(f(x)=x+1\), so

$$\lim_{x \to 2^-}f(x)=2+1=3$$

Step 2: Find the right-hand limit.

For \(x\ge 2\), \(f(x)=5\), so

$$\lim_{x \to 2^+}f(x)=5$$

Step 3: Compare them.

Since \(3\neq 5\), the two-sided limit does not exist.

$$\lim_{x \to 2}f(x) \text{ does not exist}$$

Answer: The limit does not exist.

This is called a jump discontinuity.

Worked Example 4: Behavior Near a Vertical Asymptote

Evaluate

$$\lim_{x \to 1}\frac{1}{x-1}$$

Step 1: Check the left-hand side.

If \(x\) is just less than \(1\), then \(x-1\) is a small negative number. Dividing \(1\) by a small negative number gives a very large negative value.

So

$$\lim_{x \to 1^-}\frac{1}{x-1}=-\infty$$

Step 2: Check the right-hand side.

If \(x\) is just greater than \(1\), then \(x-1\) is a small positive number. Dividing \(1\) by a small positive number gives a very large positive value.

So

$$\lim_{x \to 1^+}\frac{1}{x-1}=+\infty$$

Step 3: Compare them.

The left-hand and right-hand limits are not the same, so the two-sided limit does not exist.

$$\lim_{x \to 1}\frac{1}{x-1} \text{ does not exist}$$

Answer: The limit does not exist.

The line \(x=1\) is a vertical asymptote.

8. Important Observations About Limits

• A limit is about values near a point, not always at the point.
• A function can have a limit even if it is undefined at that point.
• If left-hand and right-hand limits are different, the limit does not exist.
• If a function is continuous at a point, the limit equals the function value there.
• Infinite behavior near a point often signals a vertical asymptote.

9. Quick Check for Evaluating a Limit

When solving a limit problem, ask yourself these questions:

1. Can I substitute directly?
2. If I get \(\frac{0}{0}\), can I factor or simplify?
3. Should I look at values in a table?
4. Do I need to check the left-hand and right-hand limits?
5. Is there a hole, jump, or vertical asymptote?

10. Why Limits Matter

Limits are the starting point for calculus because they help us describe change and behavior very precisely.

They are used to define:

• continuity,
• instantaneous rate of change,
• derivatives,
• areas and integrals later in calculus.

So when you understand limits, you are building the foundation for the rest of calculus.

Brief Summary

A limit tells us what value a function approaches as \(x\) gets close to a certain number. We can study limits using tables, graphs, direct substitution, algebraic simplification, and one-sided limits.

If both sides approach the same value, the limit exists. If they approach different values, or behave very differently near an asymptote, the limit may not exist. Limits are essential because they lead directly to continuity and the main ideas of calculus.

Formalizing Limits and One-Sided Limits

In calculus, a limit tells us what value a function is getting close to as the input gets close to a certain number. The function does not have to actually equal that value at the point. In fact, the function may not even be defined there.

This idea is especially important near boundaries, breaks, and asymptotes, where the behavior of a function can change depending on the direction from which you approach.

In this lesson, you will learn how to:

• understand the meaning of a limit,
• use one-sided limits from the left and right,
• decide whether a two-sided limit exists,
• evaluate limits using algebra and graphs,
• connect limits to continuity.

1. What is a limit?

Suppose we write

$$\lim_{x \to a} f(x) = L$$

This means that as the input values of \(x\) get closer and closer to \(a\), the output values \(f(x)\) get closer and closer to \(L\).

The key idea is approaching, not necessarily arriving. A limit looks at the behavior of the function near \(x=a\), not just at the single point \(x=a\).

For example, if a graph has a hole at \(x=2\) but the y-values on both sides are getting close to 5, then

$$\lim_{x \to 2} f(x)=5$$

even if \(f(2)\) is not 5, or even if \(f(2)\) does not exist.

2. One-sided limits

Sometimes the behavior from the left side and the right side is different. That is why we define one-sided limits.

The left-hand limit is written

$$\lim_{x \to a^-} f(x)$$

This means \(x\) approaches \(a\) using values less than \(a\), or from the left on the number line.

The right-hand limit is written

$$\lim_{x \to a^+} f(x)$$

This means \(x\) approaches \(a\) using values greater than \(a\), or from the right on the number line.

These directional limits are very important for piecewise functions, jump discontinuities, and vertical asymptotes.

3. When does the two-sided limit exist?

The two-sided limit

$$\lim_{x \to a} f(x)$$

exists only if both one-sided limits exist and are equal.

In symbols:

$$\lim_{x \to a} f(x)=L \quad \text{if and only if} \quad \lim_{x \to a^-} f(x)=L \text{ and } \lim_{x \to a^+} f(x)=L$$

If the left-hand and right-hand limits are different, then the two-sided limit does not exist.

4. Formalizing the idea of a limit

Informally, a limit means “getting close.” To make this idea precise, we say:

If we can make \(f(x)\) as close as we want to \(L\) by taking \(x\) sufficiently close to \(a\) (but not necessarily equal to \(a\)), then

$$\lim_{x \to a} f(x)=L$$

This formal idea is what makes limits reliable in calculus. We are not guessing from a graph alone. We are describing exact behavior near a point.

For one-sided limits, the same idea applies, but we approach only from one direction.

5. Reading limits from tables, graphs, and formulas

You may be asked to find a limit in different ways:

• From a graph: Look at the y-values as \(x\) approaches the target from each side.
• From a table: Check the outputs for x-values just less than and just greater than the target.
• From a formula: Use substitution first, then simplify if needed.

Always ask these questions:

1. What happens as \(x\) approaches from the left?
2. What happens as \(x\) approaches from the right?
3. Are those values equal?

6. Important cases

• Case 1: Left and right agree
  Then the two-sided limit exists.
• Case 2: Left and right are different
  Then the two-sided limit does not exist.
• Case 3: Values grow without bound
  The function may approach \(+\infty\) or \(-\infty\), often near a vertical asymptote.

For example, if

$$\lim_{x \to 3^-} f(x)=+\infty \quad \text{and} \quad \lim_{x \to 3^+} f(x)=-\infty$$

then the two-sided limit at \(x=3\) does not exist.

7. Algebraic evaluation of limits

When finding a limit from a formula, start by trying direct substitution. If substituting \(x=a\) gives a real number, that is usually the limit.

This works for functions that are continuous at that point, such as polynomials and many rational functions where the denominator is not zero.

If direct substitution gives an undefined form such as

$$\frac{0}{0}$$

then you need to simplify the expression first. Common methods include factoring and canceling common factors.

8. Limits and continuity

A function is continuous at \(x=a\) if all three of these are true:

1. \(f(a)\) exists,
2. \(\lim_{x \to a} f(x)\) exists,
3. \(\lim_{x \to a} f(x)=f(a)\).

This means the graph has no break, hole, or jump at that point.

One-sided limits help us identify when continuity fails. If the left-hand and right-hand limits are different, then the function is not continuous there.

Worked Example 1: A basic one-sided and two-sided limit

Evaluate

$$\lim_{x \to 4} (2x-1)$$

Step 1: Use direct substitution because \(2x-1\) is a polynomial and polynomials are continuous everywhere.

$$2(4)-1=8-1=7$$

So,

$$\lim_{x \to 4} (2x-1)=7$$

The left-hand limit and right-hand limit are both 7:

$$\lim_{x \to 4^-} (2x-1)=7 \quad \text{and} \quad \lim_{x \to 4^+} (2x-1)=7$$

Since they are equal, the two-sided limit exists and equals 7.

Worked Example 2: A removable discontinuity

Evaluate

$$\lim_{x \to 3} \frac{x^2-9}{x-3}$$

Step 1: Try direct substitution.

$$\frac{3^2-9}{3-3}=\frac{0}{0}$$

This is indeterminate, so simplify.

Step 2: Factor the numerator.

$$x^2-9=(x-3)(x+3)$$

So for \(x \ne 3\),

$$\frac{x^2-9}{x-3}=x+3$$

Step 3: Now take the limit of the simpler expression.

$$\lim_{x \to 3} (x+3)=6$$

Therefore,

$$\lim_{x \to 3} \frac{x^2-9}{x-3}=6$$

Even though the original function is undefined at \(x=3\), the values near 3 approach 6 from both sides, so the limit exists.

Worked Example 3: A piecewise function and one-sided limits

Let

$$f(x)=\begin{cases} 2x+1, & x<1 \\ 5, & x=1 \\ x^2+1, & x>1 \end{cases}$$

Find:

• \(\lim_{x \to 1^-} f(x)\)
• \(\lim_{x \to 1^+} f(x)\)
• \(\lim_{x \to 1} f(x)\)
• whether \(f\) is continuous at \(x=1\)

Step 1: Left-hand limit

For \(x<1\), use \(2x+1\).

$$\lim_{x \to 1^-} f(x)=2(1)+1=3$$

Step 2: Right-hand limit

For \(x>1\), use \(x^2+1\).

$$\lim_{x \to 1^+} f(x)=1^2+1=2$$

Step 3: Compare

Since

$$\lim_{x \to 1^-} f(x)=3 \quad \text{and} \quad \lim_{x \to 1^+} f(x)=2$$

the one-sided limits are not equal. Therefore,

$$\lim_{x \to 1} f(x) \text{ does not exist}$$

Step 4: Continuity

We also know \(f(1)=5\), but that does not fix the problem. Since the two-sided limit does not exist, the function is not continuous at \(x=1\).

Worked Example 4: Behavior near a vertical asymptote

Evaluate the one-sided limits and decide whether the two-sided limit exists:

$$\lim_{x \to 2^-} \frac{1}{x-2}, \qquad \lim_{x \to 2^+} \frac{1}{x-2}$$

Step 1: Approach from the left

If \(x\) is slightly less than 2, then \(x-2\) is a very small negative number. Dividing 1 by a very small negative number gives a very large negative value.

$$\lim_{x \to 2^-} \frac{1}{x-2}=-\infty$$

Step 2: Approach from the right

If \(x\) is slightly greater than 2, then \(x-2\) is a very small positive number. Dividing 1 by a very small positive number gives a very large positive value.

$$\lim_{x \to 2^+} \frac{1}{x-2}=+\infty$$

Step 3: Decide about the two-sided limit

The left and right behaviors are not the same, so

$$\lim_{x \to 2} \frac{1}{x-2}$$

does not exist.

This is a classic example of a vertical asymptote at \(x=2\).

9. A step-by-step strategy for limit problems

1. Check the type of function. Is it polynomial, rational, or piecewise?
2. Try direct substitution. If it works, you may be done.
3. If needed, simplify. Factor, cancel, or rewrite the expression.
4. For piecewise or boundary problems, find one-sided limits.
5. Compare left and right. If they match, the two-sided limit exists.
6. Check continuity if asked. Compare the limit to the function value.

10. Common mistakes to avoid

• Confusing the limit with the function value.
  \(\lim_{x \to a} f(x)\) and \(f(a)\) are not always the same.
• Ignoring one-sided behavior.
  If the function behaves differently from each side, you must check both.
• Stopping after getting \(0/0\).
  That is a sign that more algebra is needed, not the final answer.
• Saying a limit equals infinity as if it were a regular number.
  When values increase without bound, we usually say the function approaches \(+\infty\) or \(-\infty\). If left and right do not match, the two-sided limit does not exist.

11. Quick check for understanding

Ask yourself:

• Do I know the difference between \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\)?
• Can I explain why a two-sided limit exists only when the one-sided limits are equal?
• Can I simplify an expression if direct substitution gives \(0/0\)?
• Can I tell whether a function is continuous at a point?

Summary

A limit describes the value a function approaches as the input approaches a target value. A one-sided limit looks at the behavior from only one direction: left or right.

The overall two-sided limit exists only when the left-hand and right-hand limits both exist and are equal. These ideas help us analyze piecewise functions, holes, jumps, and vertical asymptotes, and they are the foundation for understanding continuity in calculus.

Algebraic Evaluation of Limits is about finding what value a function approaches as the input gets close to a certain number. In many problems, we can do this using algebra instead of making a table or graph.

This skill is important because not every function can be evaluated by simple substitution right away. Sometimes substituting gives a meaningful answer immediately, but sometimes it gives an indeterminate form like \(\frac{0}{0}\). When that happens, we use algebraic techniques such as factoring, rationalizing, or simplifying complex fractions.

In this lesson, you will learn when direct substitution works, when it does not, and how to use common algebraic methods to evaluate limits correctly.

1. What a limit means

When we write

$$\lim_{x\to a} f(x)=L$$

we mean that as \(x\) gets closer and closer to \(a\), the values of \(f(x)\) get closer and closer to \(L\).

The key idea is that a limit depends on what happens near \(x=a\), not necessarily what happens exactly at \(x=a\). In fact, the function might even be undefined at \(x=a\), and the limit can still exist.

2. Limit laws

Limit laws let us break a complicated expression into simpler parts. If

$$\lim_{x\to a} f(x)=L \quad \text{and} \quad \lim_{x\to a} g(x)=M,$$

then the following are true, as long as the expressions are defined:

• Sum law: $$\lim_{x\to a}[f(x)+g(x)] = L+M$$
• Difference law: $$\lim_{x\to a}[f(x)-g(x)] = L-M$$
• Constant multiple law: $$\lim_{x\to a}[c\,f(x)] = cL$$
• Product law: $$\lim_{x\to a}[f(x)g(x)] = LM$$
• Quotient law: $$\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M}, \quad M\ne 0$$

These laws are especially useful when direct substitution gives a real number and does not create division by zero.

3. Direct substitution

For many functions, especially polynomials and rational functions with a nonzero denominator, the easiest method is simply to substitute the value of \(x\).

If substitution gives a valid number, that number is the limit.

For example, to find

$$\lim_{x\to 2}(3x^2-5x+1),$$

substitute \(x=2\):

$$3(2)^2-5(2)+1=12-10+1=3$$

So,

$$\lim_{x\to 2}(3x^2-5x+1)=3$$

Why this works: polynomials are continuous everywhere, so their limits can be found by direct substitution.

4. When substitution fails: the indeterminate form \(\frac{0}{0}\)

Sometimes direct substitution gives

$$\frac{0}{0}$$

This does not mean the limit is 0. It means the expression needs more work. Usually, there is some algebraic simplification hiding a removable factor.

In Grade 12, the most common methods are:

• Factoring
• Rationalizing
• Simplifying a complex fraction

5. Method 1: Factoring

If substitution gives \(\frac{0}{0}\), look for a common factor in the numerator and denominator. After factoring, cancel the common factor, then substitute again.

This works because the original expression and the simplified expression are equal for all nearby values of \(x\), even if the original is undefined at one point.

Worked Example 1: A basic factoring limit

Evaluate

$$\lim_{x\to 3}\frac{x^2-9}{x-3}$$

Step 1: Try direct substitution.

$$\frac{3^2-9}{3-3}=\frac{0}{0}$$

This is indeterminate, so we factor.

Step 2: Factor the numerator.

$$x^2-9=(x-3)(x+3)$$

So the expression becomes

$$\frac{(x-3)(x+3)}{x-3}$$

Step 3: Cancel the common factor.

$$\frac{(x-3)(x+3)}{x-3}=x+3 \quad (x\ne 3)$$

Step 4: Substitute again.

$$\lim_{x\to 3}(x+3)=6$$

Answer:

$$\lim_{x\to 3}\frac{x^2-9}{x-3}=6$$

Important note: We are not plugging into the original expression after cancellation at \(x=3\). We are finding what the function approaches near \(x=3\).

Worked Example 2: A slightly harder factoring problem

Evaluate

$$\lim_{x\to 2}\frac{x^2+x-6}{x-2}$$

Step 1: Substitute.

$$\frac{2^2+2-6}{2-2}=\frac{0}{0}$$

So we factor the numerator.

Step 2: Factor.

$$x^2+x-6=(x+3)(x-2)$$

Then

$$\frac{x^2+x-6}{x-2}=\frac{(x+3)(x-2)}{x-2}$$

Step 3: Cancel.

$$\frac{(x+3)(x-2)}{x-2}=x+3 \quad (x\ne 2)$$

Step 4: Evaluate the limit.

$$\lim_{x\to 2}(x+3)=5$$

Answer:

$$\lim_{x\to 2}\frac{x^2+x-6}{x-2}=5$$

6. Method 2: Rationalizing

Rationalizing is useful when square roots cause \(\frac{0}{0}\). We remove the root from the numerator or denominator by multiplying by the conjugate.

The conjugate changes the sign between two terms:

• Conjugate of \(\sqrt{x}+2\) is \(\sqrt{x}-2\)
• Conjugate of \(3-\sqrt{x}\) is \(3+\sqrt{x}\)

The key identity is

$$ (a+b)(a-b)=a^2-b^2 $$

Worked Example 3: Rationalizing a square-root expression

Evaluate

$$\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}$$

Step 1: Substitute.

$$\frac{\sqrt{4}-2}{4-4}=\frac{0}{0}$$

We must rationalize the numerator.

Step 2: Multiply by the conjugate.

$$\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}$$

This gives

$$\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}$$

Step 3: Simplify the numerator.

$$ (\sqrt{x}-2)(\sqrt{x}+2)=x-4 $$

So the expression becomes

$$\frac{x-4}{(x-4)(\sqrt{x}+2)}$$

Step 4: Cancel the common factor.

$$\frac{x-4}{(x-4)(\sqrt{x}+2)}=\frac{1}{\sqrt{x}+2} \quad (x\ne 4)$$

Step 5: Substitute again.

$$\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{2+2}=\frac{1}{4}$$

Answer:

$$\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\frac{1}{4}$$

7. Method 3: Simplifying a complex fraction

A complex fraction is a fraction that contains fractions inside it. These often look difficult, but they can usually be simplified by combining terms or multiplying by the least common denominator.

Worked Example 4: A complex fraction limit

Evaluate

$$\lim_{x\to 1}\frac{\frac{1}{x}-1}{x-1}$$

Step 1: Substitute.

$$\frac{\frac{1}{1}-1}{1-1}=\frac{0}{0}$$

So we simplify the numerator.

Step 2: Rewrite the numerator as a single fraction.

$$\frac{1}{x}-1=\frac{1}{x}-\frac{x}{x}=\frac{1-x}{x}$$

Now the whole expression becomes

$$\frac{\frac{1-x}{x}}{x-1}$$

Step 3: Turn division into multiplication.

$$\frac{\frac{1-x}{x}}{x-1}=\frac{1-x}{x(x-1)}$$

Since \(1-x=-(x-1)\), we get

$$\frac{1-x}{x(x-1)}=\frac{-(x-1)}{x(x-1)}=-\frac{1}{x} \quad (x\ne 1)$$

Step 4: Evaluate the limit.

$$\lim_{x\to 1}\left(-\frac{1}{x}\right)=-1$$

Answer:

$$\lim_{x\to 1}\frac{\frac{1}{x}-1}{x-1}=-1$$

8. A step-by-step strategy for algebraic limit problems

1. Try direct substitution first.
2. If substitution gives a real number, that is the limit.
3. If substitution gives \(\frac{0}{0}\), simplify the expression.
4. Ask yourself which algebraic tool fits best:
   • If you see a polynomial expression, try factoring.
   • If you see a square root, try rationalizing.
   • If you see fractions inside a fraction, simplify the complex fraction.
5. After simplifying, substitute again.

9. Common mistakes to avoid

• Do not stop at \(\frac{0}{0}\). This is not the answer. It means more algebra is needed.
• Do not cancel terms unless they are factors. For example, in \(\frac{x+2}{x}\), you cannot cancel the \(x\) from just one term.
• Be careful when factoring. A small factoring mistake changes the whole answer.
• Use the conjugate correctly. Only the sign in the middle changes.
• Substitute only after simplifying. If you substitute too early in an indeterminate form, you will get stuck.

10. Quick check: when is direct substitution enough?

Direct substitution is usually enough when:

• the function is a polynomial,
• the function is a rational expression with denominator not equal to 0 at the given value,
• the expression simplifies naturally without creating \(\frac{0}{0}\).

If substitution creates a denominator of 0 and the numerator is also 0, use algebraic simplification.

11. Final summary

To evaluate limits algebraically, always begin with direct substitution. If it gives a valid number, you are done.

If substitution gives the indeterminate form \(\frac{0}{0}\), the limit may still exist. Then use algebraic methods such as factoring, rationalizing, or simplifying a complex fraction to rewrite the expression in a simpler form.

After simplification, substitute again to find the limit. With practice, you will quickly recognize which method to use and solve these limits with confidence.

Limits Involving Infinity and Asymptotes

In many functions, the most interesting behavior happens near the “edges” of the graph. Sometimes a function grows without bound, sometimes it levels off toward a fixed value, and sometimes it is not even defined at certain points. Limits involving infinity help us describe this behavior precisely.

This lesson focuses on how to analyze functions near very large positive or negative inputs, and near values where the function becomes unbounded. These ideas are closely connected to horizontal asymptotes and vertical asymptotes.

By the end of this lesson, you should be able to:

• interpret limits as \(x \to \infty\) and \(x \to -\infty\),
• interpret limits where \(f(x) \to \infty\) or \(f(x) \to -\infty\),
• identify horizontal and vertical asymptotes using limits,
• use algebra to evaluate limits involving infinity, and
• decide whether growth is bounded or unbounded.

1. What does a limit involving infinity mean?

There are two common types of limits involving infinity:

• Limits as \(x\) goes to infinity or negative infinity: these describe what happens to the function for very large positive or negative values of \(x\).
• Limits where the function goes to infinity or negative infinity: these describe what happens when the output becomes very large in magnitude near a certain \(x\)-value.

For example, in

$$\lim_{x \to \infty} f(x) = 3,$$

the input \(x\) gets larger and larger, and the output gets closer and closer to \(3\).

In

$$\lim_{x \to 2} f(x) = \infty,$$

the input \(x\) gets close to \(2\), and the output increases without bound. This does not mean the limit is a real number. It means the function becomes arbitrarily large.

2. Horizontal asymptotes

A horizontal asymptote is a horizontal line that the graph approaches as \(x\) becomes very large positive or very large negative.

Formally, if

$$\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L,$$

then the line

$$y = L$$

is a horizontal asymptote.

A function can have:

• one horizontal asymptote for both directions,
• different horizontal asymptotes as \(x \to \infty\) and \(x \to -\infty\), or
• no horizontal asymptote at all.

Example: If

$$\lim_{x \to \infty} \frac{2x+1}{x-5} = 2,$$

then \(y=2\) is a horizontal asymptote.

Important: A graph can cross a horizontal asymptote. A horizontal asymptote describes end behavior, not a barrier the graph can never pass through.

3. Vertical asymptotes

A vertical asymptote occurs when the function grows without bound near some \(x\)-value.

Formally, if at least one of the following is true:

$$\lim_{x \to a^-} f(x) = \infty, \quad \lim_{x \to a^-} f(x) = -\infty, \quad \lim_{x \to a^+} f(x) = \infty, \quad \lim_{x \to a^+} f(x) = -\infty,$$

then

$$x = a$$

is a vertical asymptote.

The one-sided limits matter. Sometimes the function goes to \(\infty\) on one side and \(-\infty\) on the other side. It is still a vertical asymptote.

4. Bounded vs unbounded growth

A function is showing bounded behavior if its outputs stay near a finite value. For example, if

$$\lim_{x \to \infty} f(x) = 4,$$

then as \(x\) gets large, the function is bounded near \(4\).

A function shows unbounded growth if the outputs increase without limit or decrease without limit. For example:

• $$\lim_{x \to 3^-} f(x) = \infty$$
• $$\lim_{x \to \infty} f(x) = \infty$$

In both cases, the function is not approaching a finite number.

5. How to evaluate limits as \(x \to \infty\) or \(x \to -\infty\)

Different kinds of functions are handled in different ways.

A. Rational functions

A rational function is a quotient of polynomials, such as

$$f(x)=\frac{3x^2+1}{x^2-4x+7}.$$

For rational functions, compare the degrees of the numerator and denominator.

• If the numerator has lower degree than the denominator, then the limit is \(0\).
• If the degrees are equal, then the limit is the ratio of the leading coefficients.
• If the numerator has higher degree, then there is no horizontal asymptote; the function usually grows without bound or decreases without bound.

This works because the highest-power terms dominate when \(x\) becomes very large.

B. Dividing by the highest power

A reliable method is to divide every term in the fraction by the highest power of \(x\) in the denominator or numerator. This makes the end behavior easier to see.

For example,

$$\frac{2x+1}{x-5} = \frac{2 + \frac{1}{x}}{1 - \frac{5}{x}}.$$

As \(x \to \infty\), both \(\frac{1}{x}\) and \(\frac{5}{x}\) go to \(0\), so the limit becomes

$$\frac{2+0}{1-0}=2.$$

C. Looking near a possible vertical asymptote

To study a vertical asymptote, factor the denominator if possible and check the sign of the expression from the left and right sides.

This is important because the sign determines whether the function approaches \(\infty\) or \(-\infty\).

6. Formal meaning of asymptotes using limits

It is important to connect asymptotes to limit language.

• \(y=L\) is a horizontal asymptote if \(\lim_{x \to \infty} f(x)=L\) or \(\lim_{x \to -\infty} f(x)=L\).
• \(x=a\) is a vertical asymptote if \(f(x)\) becomes unbounded as \(x\) approaches \(a\) from one side or both sides.

So asymptotes are not just guessed from graphs. They are justified by limit calculations.

7. Worked Examples

Example 1: Finding a horizontal asymptote

Evaluate

$$\lim_{x \to \infty} \frac{5x-2}{2x+3}.$$

Step 1: Notice that the numerator and denominator both have degree 1.

Step 2: For equal degrees, the limit is the ratio of the leading coefficients.

$$\lim_{x \to \infty} \frac{5x-2}{2x+3} = \frac{5}{2}.$$

Answer: The limit is \(\frac{5}{2}\), so the horizontal asymptote is

$$y=\frac{5}{2}.$$

Why this makes sense: For very large \(x\), the terms \(-2\) and \(+3\) matter very little compared to \(5x\) and \(2x\).

Example 2: End behavior with different degrees

Evaluate

$$\lim_{x \to \infty} \frac{3x^2+4}{x^3-1}.$$

Step 1: Compare degrees.

• Numerator degree: 2
• Denominator degree: 3

The denominator has the higher degree, so the fraction should approach \(0\).

Step 2: Confirm by dividing by \(x^3\):

$$\frac{3x^2+4}{x^3-1} = \frac{\frac{3}{x}+\frac{4}{x^3}}{1-\frac{1}{x^3}}.$$

As \(x \to \infty\), each term with \(\frac{1}{x}\) goes to \(0\). So:

$$\lim_{x \to \infty} \frac{3x^2+4}{x^3-1} = \frac{0+0}{1-0}=0.$$

Answer: The limit is \(0\), so the horizontal asymptote is

$$y=0.$$

Example 3: Finding a vertical asymptote with one-sided limits

Analyze the behavior of

$$f(x)=\frac{1}{x-2}$$

near \(x=2\).

Step 1: Consider the left-hand limit.

If \(x\) is just less than \(2\), then \(x-2\) is a very small negative number.

So

$$\lim_{x \to 2^-} \frac{1}{x-2} = -\infty.$$

Step 2: Consider the right-hand limit.

If \(x\) is just greater than \(2\), then \(x-2\) is a very small positive number.

So

$$\lim_{x \to 2^+} \frac{1}{x-2} = \infty.$$

Conclusion: Since the function becomes unbounded near \(x=2\), the line

$$x=2$$

is a vertical asymptote.

Example 4: A more detailed rational function

Find all vertical and horizontal asymptotes of

$$f(x)=\frac{2x+1}{x^2-9}.$$

Step 1: Factor the denominator.

$$x^2-9=(x-3)(x+3).$$

The denominator is zero at \(x=3\) and \(x=-3\). These are possible vertical asymptotes.

Step 2: Check whether the numerator is also zero there.

• At \(x=3\), numerator \(=2(3)+1=7 \neq 0\)
• At \(x=-3\), numerator \(=2(-3)+1=-5 \neq 0\)

So both values give vertical asymptotes.

Vertical asymptotes:

$$x=3 \quad \text{and} \quad x=-3.$$

Step 3: Find the horizontal asymptote.

The numerator has degree 1 and the denominator has degree 2. Since the denominator has higher degree,

$$\lim_{x \to \infty} \frac{2x+1}{x^2-9}=0$$

and

$$\lim_{x \to -\infty} \frac{2x+1}{x^2-9}=0.$$

So the horizontal asymptote is

$$y=0.$$

8. Common mistakes to avoid

• Confusing \(x \to \infty\) with \(f(x) \to \infty\). One describes the input; the other describes the output.
• Forgetting one-sided limits when checking vertical asymptotes.
• Assuming every undefined point is a vertical asymptote. A function can be undefined at a point without becoming unbounded there.
• Using only the graph without algebra. Limit calculations give the formal justification.
• Thinking a horizontal asymptote cannot be crossed. It can be crossed because it only describes end behavior.

9. Quick strategy guide

1. If the question asks about behavior as \(x \to \infty\) or \(x \to -\infty\), look at the highest-power terms.
2. For rational functions, compare degrees or divide by the highest power.
3. If the question asks about a possible vertical asymptote, find where the denominator is zero.
4. Check left-hand and right-hand behavior separately.
5. Use limit notation to state the asymptote clearly.

10. Brief Summary

Limits involving infinity help describe end behavior and unbounded behavior of functions. If \(f(x)\) approaches a finite value as \(x\) becomes very large or very negative, that value gives a horizontal asymptote. If \(f(x)\) becomes infinitely large in magnitude near \(x=a\), then \(x=a\) is a vertical asymptote.

For rational functions, end behavior is usually determined by comparing degrees of the numerator and denominator. Vertical asymptotes are often found by locating where the denominator is zero and checking one-sided limits. These ideas let us describe clearly whether a function stays bounded or grows without bound.

Special Trigonometric Limits are some of the most important limits in calculus. They help us evaluate expressions involving sine, cosine, and tangent near zero, especially when direct substitution gives an indeterminate form like \(\frac{0}{0}\).

In this lesson, you will learn how to derive the key trigonometric limit using the Squeeze Theorem, and then how to use it to evaluate many related limits. This idea is a foundation for later topics such as derivatives of trigonometric functions.

Main goal: understand and use the limit

$$\lim_{x\to 0} \frac{\sin x}{x} = 1$$

and related results such as

$$\lim_{x\to 0} \frac{\tan x}{x} = 1 \qquad \text{and} \qquad \lim_{x\to 0} \frac{1-\cos x}{x} = 0.$$

Throughout this lesson, angles are measured in radians. This is very important. These special limits are true in this form only when \(x\) is in radians.

Why radians matter: if angles are measured in degrees, then the limit \(\frac{\sin x}{x}\) as \(x \to 0\) is not 1. So whenever you see these standard trigonometric limits in calculus, assume radians unless stated otherwise.

1. Review: What is a limit?

When we write \(\lim_{x\to a} f(x)\), we are asking: what value does \(f(x)\) get close to as \(x\) gets close to \(a\)? The function does not even need to be defined at \(x=a\) for the limit to exist.

For example, if we try direct substitution in

$$\lim_{x\to 0} \frac{\sin x}{x},$$

we get

$$\frac{\sin 0}{0} = \frac{0}{0},$$

which is indeterminate. That means direct substitution does not work, so we need another method.

2. The Squeeze Theorem

The Squeeze Theorem says that if

$$g(x) \le f(x) \le h(x)$$

for all \(x\) near \(a\), and if

$$\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L,$$

then

$$\lim_{x\to a} f(x) = L.$$

In simple words: if a function is trapped between two other functions that both approach the same value, then the middle function must approach that value too.

This theorem is the key to proving the most famous special trigonometric limit.

3. Deriving \(\lim_{x\to 0} \frac{\sin x}{x} = 1\)

Consider a circle of radius 1, called the unit circle. For a small positive angle \(x\) in radians, geometry gives the inequality

$$\sin x < x < \tan x \qquad \text{for } 0 < x < \frac{\pi}{2}.$$

This inequality comes from comparing areas or lengths inside the unit circle. The important result is the inequality itself.

Now divide every part by \(\sin x\), which is positive for \(0 $$1 < \frac{x}{\sin x} < \frac{\tan x}{\sin x}.$$

Since \(\tan x = \frac{\sin x}{\cos x}\), we get

$$\frac{\tan x}{\sin x} = \frac{1}{\cos x}.$$

So the inequality becomes

$$1 < \frac{x}{\sin x} < \frac{1}{\cos x}.$$

Now take reciprocals. Since all quantities are positive, the inequality reverses:

$$\cos x < \frac{\sin x}{x} < 1.$$

As \(x \to 0\), we know \(\cos x \to 1\). So we have

$$\cos x \to 1 \qquad \text{and} \qquad 1 \to 1.$$

By the Squeeze Theorem,

$$\lim_{x\to 0} \frac{\sin x}{x} = 1.$$

This is one of the most important limits in all of calculus.

4. A useful equivalent form

Since

$$\lim_{x\to 0} \frac{\sin x}{x} = 1,$$

we can also write

$$\lim_{x\to 0} \frac{x}{\sin x} = 1.$$

Both forms are used often. If one appears in a question, you should recognize it immediately.

5. Another key limit: \(\lim_{x\to 0} \frac{\tan x}{x} = 1\)

Start with

$$\frac{\tan x}{x} = \frac{\sin x}{x\cos x} = \left(\frac{\sin x}{x}\right)\left(\frac{1}{\cos x}\right).$$

Now as \(x \to 0\),

$$\frac{\sin x}{x} \to 1 \qquad \text{and} \qquad \cos x \to 1,$$

so

$$\frac{1}{\cos x} \to 1.$$

Therefore,

$$\lim_{x\to 0} \frac{\tan x}{x} = 1 \cdot 1 = 1.$$

6. Limits involving \(1-\cos x\)

A common expression is

$$\lim_{x\to 0} \frac{1-\cos x}{x}.$$

To evaluate it, multiply by the conjugate:

$$\frac{1-\cos x}{x} \cdot \frac{1+\cos x}{1+\cos x} = \frac{1-\cos^2 x}{x(1+\cos x)} = \frac{\sin^2 x}{x(1+\cos x)}.$$

Then rewrite:

$$\frac{\sin^2 x}{x(1+\cos x)} = \left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{1+\cos x}\right).$$

As \(x\to 0\), we know \(\frac{\sin x}{x} \to 1\), and also \(\sin x \to 0\), \(\cos x \to 1\), so

$$\frac{\sin x}{1+\cos x} \to \frac{0}{2} = 0.$$

Therefore,

$$\lim_{x\to 0} \frac{1-\cos x}{x} = 1\cdot 0 = 0.$$

There is another very important related limit:

$$\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}.$$

This one appears often in advanced problems. We can derive it using the same conjugate idea:

$$\frac{1-\cos x}{x^2} \cdot \frac{1+\cos x}{1+\cos x} = \frac{\sin^2 x}{x^2(1+\cos x)} = \frac{\left(\frac{\sin x}{x}\right)^2}{1+\cos x}.$$

Now let \(x\to 0\):

$$\frac{\left(\frac{\sin x}{x}\right)^2}{1+\cos x} \to \frac{1^2}{1+1} = \frac{1}{2}.$$

So,

$$\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}.$$

7. How to handle limits with constants and substitutions

You will often see expressions such as \(\sin(3x)\), \(\tan(5x)\), or \(\frac{\sin 7x}{x}\). In these cases, try to rewrite the expression so that the standard limit appears.

For example,

$$\frac{\sin 3x}{x} = 3\cdot \frac{\sin 3x}{3x}.$$

As \(x\to 0\), \(3x\to 0\), so

$$\lim_{x\to 0} \frac{\sin 3x}{x} = 3 \cdot 1 = 3.$$

This method is very important. Always look for a way to create \(\frac{\sin u}{u}\) or \(\frac{\tan u}{u}\).

8. Summary of standard special trigonometric limits

• $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$
• $$\lim_{x\to 0} \frac{x}{\sin x} = 1$$
• $$\lim_{x\to 0} \frac{\tan x}{x} = 1$$
• $$\lim_{x\to 0} \frac{1-\cos x}{x} = 0$$
• $$\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$$

9. Worked Examples

Example 1: Evaluate

$$\lim_{x\to 0} \frac{\sin x}{x}.$$

Solution:

This is the standard special trigonometric limit. Therefore,

$$\lim_{x\to 0} \frac{\sin x}{x} = 1.$$

Answer: \(1\)

Example 2: Evaluate

$$\lim_{x\to 0} \frac{\sin 5x}{x}.$$

Solution:

Rewrite the expression to match the standard form:

$$\frac{\sin 5x}{x} = 5\cdot \frac{\sin 5x}{5x}.$$

Now as \(x\to 0\), \(5x\to 0\), so

$$\lim_{x\to 0} \frac{\sin 5x}{5x} = 1.$$

Therefore,

$$\lim_{x\to 0} \frac{\sin 5x}{x} = 5\cdot 1 = 5.$$

Answer: \(5\)

Example 3: Evaluate

$$\lim_{x\to 0} \frac{\tan 2x}{x}.$$

Solution:

Rewrite:

$$\frac{\tan 2x}{x} = 2\cdot \frac{\tan 2x}{2x}.$$

Using the standard limit \(\lim_{u\to 0} \frac{\tan u}{u}=1\), we get

$$\lim_{x\to 0} \frac{\tan 2x}{2x} = 1.$$

So,

$$\lim_{x\to 0} \frac{\tan 2x}{x} = 2\cdot 1 = 2.$$

Answer: \(2\)

Example 4: Evaluate

$$\lim_{x\to 0} \frac{1-\cos x}{x^2}.$$

Solution:

Multiply by the conjugate:

$$\frac{1-\cos x}{x^2} \cdot \frac{1+\cos x}{1+\cos x} = \frac{1-\cos^2 x}{x^2(1+\cos x)}.$$

Use the identity \(1-\cos^2 x = \sin^2 x\):

$$= \frac{\sin^2 x}{x^2(1+\cos x)} = \frac{\left(\frac{\sin x}{x}\right)^2}{1+\cos x}.$$

Now take the limit:

$$\lim_{x\to 0} \frac{\left(\frac{\sin x}{x}\right)^2}{1+\cos x} = \frac{1^2}{2} = \frac{1}{2}.$$

Answer: \(\frac{1}{2}\)

10. Common mistakes to avoid

• Using degrees instead of radians. These standard limits require radians.
• Substituting too early. If direct substitution gives \(\frac{0}{0}\), rewrite the expression first.
• Forgetting to match the standard form. For example, in \(\frac{\sin 4x}{x}\), create \(\frac{\sin 4x}{4x}\) by multiplying by 4.
• Mixing up formulas. Remember that \(\lim_{x\to 0} \frac{1-\cos x}{x} = 0\), but \(\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}\).

11. Quick strategy for solving these problems

1. Check what happens with direct substitution.
2. If you get \(\frac{0}{0}\), look for a standard trigonometric limit.
3. Rewrite the expression to create one of these forms:
   • \(\frac{\sin u}{u}\)
   • \(\frac{\tan u}{u}\)
   • an expression involving \(1-\cos x\), often using a conjugate
4. Apply the known limit and simplify.

12. Brief Summary

The most important special trigonometric limit is

$$\lim_{x\to 0} \frac{\sin x}{x} = 1,$$

which is proved using the Squeeze Theorem. From it, we can derive other useful limits such as

$$\lim_{x\to 0} \frac{\tan x}{x} = 1 \quad \text{and} \quad \lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}.$$

These limits help us understand how trigonometric functions behave near \(x=0\), especially when direct substitution does not work. The key skill is learning how to rewrite an expression so it matches a standard limit form.

Continuity at a Point and Over Intervals is one of the main ideas in limits and calculus. Informally, a function is continuous if its graph has no breaks, holes, or jumps. But in Grade 12 Mathematics, we need a precise mathematical definition.

This lesson explains what it means for a function to be continuous at a single point and on an interval, how to test continuity, and how continuity theorems help us decide quickly whether common functions are continuous.

Understanding continuity is important because many later ideas in calculus, such as derivatives and optimization, depend on whether a function behaves smoothly near a point.

1. The formal definition of continuity at a point

A function \(f\) is continuous at a point \(x=a\) if all three of the following conditions are true:

• The function value exists: \(f(a)\) is defined.
• The limit exists: \(\lim_{x \to a} f(x)\) exists.
• The limit equals the function value: $$\lim_{x \to a} f(x)=f(a)$$

If even one of these conditions fails, then the function is not continuous at \(x=a\).

This gives us a simple checklist:

1. Can you plug in \(a\) and get a function value?
2. Do the left-hand and right-hand limits agree?
3. Does that common limit equal the actual function value?

2. Why all three conditions matter

Each condition checks a different part of the function's behavior.

• If \(f(a)\) does not exist, then there is no actual point on the graph at \(x=a\).
• If the limit does not exist, the function does not approach one single value as \(x\) gets close to \(a\).
• If the limit and function value are different, the graph may have a hole or a misplaced point.

So continuity means the function's behavior near the point matches its value at the point.

3. Left-hand and right-hand limits

The limit \(\lim_{x \to a} f(x)\) exists only if both one-sided limits exist and are equal:

$$\lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)$$

If these two are different, then the function has a jump or mismatch at \(x=a\), so it cannot be continuous there.

4. Continuity on an interval

A function is continuous on an interval if it is continuous at every point in that interval.

For an open interval such as \((1,5)\), this means continuity at every point strictly between 1 and 5.

For a closed interval such as \([1,5]\), we also need the correct endpoint behavior:

• At \(x=1\), the function must be continuous from the right: $$\lim_{x \to 1^+} f(x)=f(1)$$
• At \(x=5\), the function must be continuous from the left: $$\lim_{x \to 5^-} f(x)=f(5)$$

So continuity on a closed interval means:

• continuous at every interior point,
• right-continuous at the left endpoint,
• left-continuous at the right endpoint.

5. Common types of discontinuity

When a function is not continuous, the break usually fits one of these patterns:

• Removable discontinuity: the limit exists, but \(f(a)\) is missing or has the wrong value. This often creates a hole.
• Jump discontinuity: the left-hand and right-hand limits are different.
• Infinite discontinuity: the function grows without bound near the point, often near a vertical asymptote.

6. Functions that are continuous automatically

Many common functions are continuous wherever they are defined. This lets us use continuity theorems instead of checking the full definition every time.

• Polynomials are continuous for all real numbers.
• Rational functions are continuous wherever the denominator is not zero.
• Root functions are continuous wherever they are defined.
• Absolute value functions are continuous for all real numbers.
• Sums, differences, products, and quotients of continuous functions are continuous, as long as the quotient does not divide by zero.

These facts are called continuity theorems. They are very useful when identifying intervals of continuity.

7. How to test continuity at a point

Use this process:

1. Find \(f(a)\).
2. Find \(\lim_{x \to a} f(x)\).
3. Compare the two values.

If the function is piecewise, be especially careful at the boundary points where the formula changes.

Worked Example 1: A polynomial at a point

Determine whether \(f(x)=x^2-3x+2\) is continuous at \(x=1\).

Step 1: Find the function value.

$$f(1)=1^2-3(1)+2=1-3+2=0$$

Step 2: Find the limit.

Since \(f(x)\) is a polynomial, it is continuous for all real numbers. So:

$$\lim_{x \to 1} (x^2-3x+2)=f(1)=0$$

Step 3: Compare.

$$\lim_{x \to 1} f(x)=0=f(1)$$

Conclusion: \(f\) is continuous at \(x=1\).

This is a simple example where the theorem about polynomials saves time.

Worked Example 2: A removable discontinuity

Determine whether the function $$f(x)=\begin{cases} \dfrac{x^2-4}{x-2}, & x\ne 2 \\ 5, & x=2 \end{cases}$$ is continuous at \(x=2\).

Step 1: Find the function value.

$$f(2)=5$$

Step 2: Find the limit as \(x\to 2\).

For \(x\ne 2\), factor the numerator:

$$\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$$

So near \(x=2\), the function behaves like \(x+2\). Therefore:

$$\lim_{x \to 2} \frac{x^2-4}{x-2}=\lim_{x \to 2}(x+2)=4$$

Step 3: Compare the limit and the function value.

$$\lim_{x \to 2} f(x)=4 \quad \text{but} \quad f(2)=5$$

Since these are not equal, the function is not continuous at \(x=2\).

Conclusion: This is a removable discontinuity. If the function had been defined with \(f(2)=4\), then it would be continuous at \(x=2\).

Worked Example 3: A piecewise function with a possible jump

Check whether $$f(x)=\begin{cases} 2x+1, & x<3 \\ x^2-4, & x\ge 3 \end{cases}$$ is continuous at \(x=3\).

Step 1: Find the function value.

Since the second rule applies at \(x=3\):

$$f(3)=3^2-4=9-4=5$$

Step 2: Find the left-hand limit.

$$\lim_{x \to 3^-} f(x)=\lim_{x \to 3^-}(2x+1)=2(3)+1=7$$

Step 3: Find the right-hand limit.

$$\lim_{x \to 3^+} f(x)=\lim_{x \to 3^+}(x^2-4)=3^2-4=5$$

Step 4: Compare the one-sided limits.

$$\lim_{x \to 3^-} f(x)=7 \quad \text{and} \quad \lim_{x \to 3^+} f(x)=5$$

Since the left-hand and right-hand limits are not equal, the two-sided limit does not exist.

Conclusion: \(f\) is not continuous at \(x=3\). This is a jump discontinuity.

Worked Example 4: Continuity over intervals for a rational function

Find the intervals on which $$f(x)=\frac{x+1}{x^2-9}$$ is continuous.

Step 1: Identify the type of function.

This is a rational function, so it is continuous wherever its denominator is not zero.

Step 2: Find where the denominator is zero.

$$x^2-9=0$$ $$x^2=9$$ $$x=\pm 3$$

So the function is undefined at \(x=-3\) and \(x=3\).

Step 3: Write the intervals.

The function is continuous on:

$$(-\infty,-3), \quad (-3,3), \quad (3,\infty)$$

Conclusion: Rational functions are continuous on each interval of their domain. The points \(x=-3\) and \(x=3\) are breaks in continuity and correspond to vertical asymptotes.

8. Continuity and direct substitution

If a function is known to be continuous at \(x=a\), then limits are easier to evaluate because:

$$\lim_{x \to a} f(x)=f(a)$$

This means you can often use direct substitution. For example, since \(\sqrt{x+5}\) is continuous where defined,

$$\lim_{x \to 4} \sqrt{x+5}=\sqrt{4+5}=3$$

But direct substitution only works when the function is continuous at that point.

9. Important caution near boundaries and endpoints

When working on restricted domains, such as \([0,\infty)\), the function may still be continuous at an endpoint even though you can only approach from one side.

For example, \(f(x)=\sqrt{x}\) is continuous on \([0,\infty)\). At \(x=0\), we use the right-hand limit:

$$\lim_{x \to 0^+} \sqrt{x}=0=f(0)$$

So endpoint continuity uses the limit from within the interval.

10. Common mistakes students make

• Thinking that a function is continuous just because \(f(a)\) exists.
• Forgetting to check whether the limit exists.
• Ignoring one-sided limits in piecewise functions.
• Using direct substitution when the function is not continuous there.
• Confusing “undefined” with “limit does not exist.” These are different ideas.

11. Quick checklist for exams

When asked whether a function is continuous at \(x=a\), ask:

1. Is \(f(a)\) defined?
2. Does \(\lim_{x \to a} f(x)\) exist?
3. Is \(\lim_{x \to a} f(x)=f(a)\)?

When asked where a function is continuous, ask:

1. What type of function is it?
2. Where is it defined?
3. Are there points where the rule changes or the denominator becomes zero?
4. Write the intervals between those points.

Brief Summary

A function is continuous at a point \(x=a\) if the function value exists, the limit exists, and the two are equal. A function is continuous on an interval if it is continuous at every point of that interval, with one-sided continuity used at endpoints of closed intervals.

Polynomials, rational functions, roots, and many other familiar functions are continuous wherever they are defined. To check continuity, always use the three-part definition and be especially careful with piecewise functions, holes, jumps, and vertical asymptotes.

Classification of Discontinuities

In calculus, a function is continuous at a point if its graph has no break there. Informally, you can draw the graph near that point without lifting your pencil. More formally, a function \(f(x)\) is continuous at \(x=a\) when all three of these are true:

• \(f(a)\) is defined,
• \(\lim_{x\to a} f(x)\) exists,
• \(\lim_{x\to a} f(x)=f(a)\).

If any one of these conditions fails, the function is discontinuous at \(x=a\). In this lesson, we will classify discontinuities into three main types:

• Removable discontinuity — a hole in the graph,
• Jump discontinuity — the left and right sides approach different values,
• Infinite discontinuity — the function grows without bound near a point, creating a vertical asymptote.

Learning to classify discontinuities is important because it helps you understand a function’s behavior near boundaries and asymptotes, and it connects directly to limits.

1. Removable Discontinuity

A removable discontinuity happens when the limit exists at a point, but the function is either not defined there or is defined incorrectly. This creates a hole in the graph.

In symbols, at \(x=a\), a removable discontinuity occurs when:

$$\lim_{x\to a} f(x) = L$$

exists, but either:

• \(f(a)\) does not exist, or
• \(f(a)\neq L\).

This kind of discontinuity is called “removable” because if we redefine the function value at that point to equal the limit, the function becomes continuous there.

A common place to see this is in rational functions where a factor cancels.

2. Jump Discontinuity

A jump discontinuity happens when the left-hand and right-hand limits both exist, but they are not equal.

At \(x=a\):

$$\lim_{x\to a^-} f(x) \neq \lim_{x\to a^+} f(x)$$

Since the two one-sided limits are different, the overall limit does not exist. On the graph, this looks like a sudden step or jump.

This often appears in piecewise-defined functions.

3. Infinite Discontinuity

An infinite discontinuity happens when the function increases or decreases without bound as \(x\) approaches a certain value. This usually means there is a vertical asymptote.

For example, if as \(x\to a\),

$$f(x)\to \infty \quad \text{or} \quad f(x)\to -\infty,$$

then \(x=a\) is a point of infinite discontinuity.

The limit does not exist as a finite number, but the behavior tells us the function is approaching a vertical asymptote.

How to Classify a Discontinuity

When you are given a function and asked to classify a discontinuity at \(x=a\), follow these steps:

1. Check whether \(f(a)\) is defined.
2. Find the left-hand limit \(\lim_{x\to a^-} f(x)\).
3. Find the right-hand limit \(\lim_{x\to a^+} f(x)\).
4. Compare the one-sided limits.
5. Use what you find to classify the discontinuity.

You can use this quick guide:

• If the left and right limits are equal and finite, but the function value is missing or different, it is removable.
• If the left and right limits are finite but unequal, it is a jump.
• If one or both sides go to \(\infty\) or \(-\infty\), it is infinite.

Worked Example 1: Removable Discontinuity

Classify the discontinuity of

$$f(x)=\frac{x^2-1}{x-1}$$

at \(x=1\).

Step 1: Check whether the function is defined at \(x=1\).

Substitute \(x=1\):

$$f(1)=\frac{1^2-1}{1-1}=\frac{0}{0}$$

This is undefined, so \(f(1)\) does not exist.

Step 2: Simplify the function.

$$x^2-1=(x-1)(x+1)$$

So for \(x\neq 1\),

$$f(x)=\frac{(x-1)(x+1)}{x-1}=x+1$$

Step 3: Find the limit as \(x\to 1\).

$$\lim_{x\to 1} f(x)=\lim_{x\to 1}(x+1)=2$$

The limit exists and is finite, but the function is not defined at \(x=1\).

Conclusion: The discontinuity at \(x=1\) is removable.

The graph is the line \(y=x+1\) with a hole at \((1,2)\).

Worked Example 2: Jump Discontinuity

Classify the discontinuity of

$$f(x)= \begin{cases} 2x+1, & x<3 \\ 7, & x\ge 3 \end{cases}$$

at \(x=3\).

Step 1: Find the left-hand limit.

For \(x<3\), \(f(x)=2x+1\), so

$$\lim_{x\to 3^-} f(x)=2(3)+1=7$$

Step 2: Find the right-hand limit.

For \(x\ge 3\), \(f(x)=7\), so

$$\lim_{x\to 3^+} f(x)=7$$

Since both one-sided limits are equal to 7, this function is actually continuous at \(x=3\), because \(f(3)=7\) as well.

So this is not a discontinuity.

Let us slightly change the example to create a jump:

$$g(x)= \begin{cases} 2x+1, & x<3 \\ 8, & x\ge 3 \end{cases}$$

Now:

$$\lim_{x\to 3^-} g(x)=7$$ $$\lim_{x\to 3^+} g(x)=8$$

The one-sided limits are different, so the limit at \(x=3\) does not exist.

Conclusion: \(g(x)\) has a jump discontinuity at \(x=3\).

Worked Example 3: Infinite Discontinuity

Classify the discontinuity of

$$f(x)=\frac{1}{x-4}$$

at \(x=4\).

Step 1: Check whether the function is defined at \(x=4\).

$$f(4)=\frac{1}{4-4}=\frac{1}{0}$$

This is undefined.

Step 2: Examine the one-sided limits.

As \(x\to 4^-\), the denominator \(x-4\) is a small negative number, so

$$\lim_{x\to 4^-} \frac{1}{x-4}=-\infty$$

As \(x\to 4^+\), the denominator \(x-4\) is a small positive number, so

$$\lim_{x\to 4^+} \frac{1}{x-4}=\infty$$

The function does not approach a finite value. Instead, it grows without bound on both sides.

Conclusion: There is an infinite discontinuity at \(x=4\), and \(x=4\) is a vertical asymptote.

Worked Example 4: Choosing the Correct Type Carefully

Classify the discontinuity of

$$f(x)=\frac{x^2-4}{x-2}$$

at \(x=2\).

At first glance, since the denominator is 0 at \(x=2\), you might think there is a vertical asymptote. But we must simplify first.

Factor the numerator:

$$x^2-4=(x-2)(x+2)$$

So for \(x\neq 2\),

$$f(x)=x+2$$

Now find the limit:

$$\lim_{x\to 2} f(x)=\lim_{x\to 2}(x+2)=4$$

But \(f(2)\) is undefined in the original function because the denominator becomes 0.

Conclusion: The discontinuity is removable, not infinite.

This example shows an important idea: a zero in the denominator does not always mean a vertical asymptote. If the factor cancels and the limit exists, the discontinuity is removable.

Common Mistakes to Avoid

• Confusing removable and infinite discontinuities: If a factor cancels and leaves a finite limit, it is removable.
• Forgetting one-sided limits: Jump discontinuities can only be identified by comparing the left and right limits.
• Looking only at whether the function is undefined: Many discontinuities are undefined, but the type depends on the limit behavior.
• Assuming every piecewise function has a jump: Some piecewise functions are actually continuous at the joining point.

Quick Comparison Table

• Removable: limit exists, but function value is missing or wrong; graph has a hole.
• Jump: left and right limits exist but are different; graph has a step.
• Infinite: function goes to \(\infty\) or \(-\infty\); graph has a vertical asymptote.

How This Connects to Limits and Continuity

Discontinuities are really about what happens to a function near a point. That is why limits are the main tool for classifying them.

To decide whether a function is continuous, you do not only check the value at the point. You also study the behavior from the left and from the right. This is especially important near boundaries, joins in piecewise functions, and vertical asymptotes.

Summary

A function is continuous at \(x=a\) when the function value exists, the limit exists, and they are equal. If not, the function is discontinuous there.

The three main types are:

• Removable discontinuity: the limit exists, but there is a hole.
• Jump discontinuity: the left and right limits are different.
• Infinite discontinuity: the function becomes unbounded near the point, giving a vertical asymptote.

When classifying a discontinuity, always check the function value, then the left-hand and right-hand limits. The limit behavior tells you the correct type.

Intermediate Value Theorem (IVT) is one of the most important results about continuous functions. It tells us that if a function is continuous on an interval, then its graph cannot “jump” over any value between two outputs.

This theorem is especially useful for showing that an equation has a solution, even when we cannot solve it exactly. In many problems, IVT is used to guarantee the existence of a root, meaning a value of \(x\) for which \(f(x)=0\).

To understand IVT well, we need to focus on two ideas:

• Continuity: the function has no breaks, holes, or jumps on the interval.
• Intermediate values: every output between two endpoint outputs must also occur somewhere in between.

Imagine drawing a continuous curve from one point to another without lifting your pencil. If one endpoint is above the \(x\)-axis and the other is below it, the graph must cross the \(x\)-axis somewhere in between. That is the basic idea behind IVT.

The Intermediate Value Theorem can be stated formally like this:

If \(f\) is continuous on the closed interval \([a,b]\), and \(N\) is any number between \(f(a)\) and \(f(b)\), then there exists some number \(c\) in \([a,b]\) such that

$$f(c)=N.$$

In simpler words, if a continuous function starts at one output and ends at another, then it must take on every output value in between at least once.

A very common special case happens when we want to show there is a root. If \(f\) is continuous on \([a,b]\) and

$$f(a)\cdot f(b)<0,$$

then \(f(a)\) and \(f(b)\) have opposite signs. One is positive and the other is negative. Since \(0\) lies between a positive and a negative number, IVT guarantees that there is some \(c\in(a,b)\) such that

$$f(c)=0.$$

This is often called the existence of a root.

Why continuity matters is very important. IVT only works if the function is continuous on the whole interval. If there is a jump or break, the function might skip over values.

For example, suppose a function has a jump from \(-2\) to \(3\). Even though \(0\) is between \(-2\) and \(3\), the function might never actually equal \(0\). So before using IVT, always check continuity first.

How to use IVT in a problem:

1. Identify the function \(f(x)\).
2. Check that \(f\) is continuous on the interval \([a,b]\).
3. Evaluate \(f(a)\) and \(f(b)\).
4. Check whether the target value \(N\) lies between \(f(a)\) and \(f(b)\).
5. Conclude that there exists some \(c\in[a,b]\) such that \(f(c)=N\).

When the target value is \(0\), step 4 usually becomes checking whether \(f(a)\) and \(f(b)\) have opposite signs.

Worked Example 1: Showing a root exists

Show that the equation

$$x^3-x-2=0$$

has at least one solution between \(1\) and \(2\).

Step 1: Define the function.

$$f(x)=x^3-x-2$$

Step 2: Check continuity. This is a polynomial, and all polynomials are continuous for every real number. So \(f\) is continuous on \([1,2]\).

Step 3: Evaluate the endpoints.

$$f(1)=1^3-1-2=-2$$ $$f(2)=2^3-2-2=4$$

Step 4: Look at the signs. Since \(f(1)=-2\) and \(f(2)=4\), one value is negative and the other is positive. So

$$f(1)\cdot f(2)=(-2)(4)<0.$$

Conclusion: By the Intermediate Value Theorem, there exists some \(c\in(1,2)\) such that

$$f(c)=0.$$

So the equation \(x^3-x-2=0\) has at least one real solution between \(1\) and \(2\).

Notice that IVT does not tell us the exact solution. It only guarantees that a solution exists.

Worked Example 2: Showing a function reaches a specific value

Show that the function

$$f(x)=x^2+1$$

takes the value \(3\) somewhere on the interval \([1,2]\).

Step 1: Continuity. Since \(f(x)=x^2+1\) is a polynomial, it is continuous on \([1,2]\).

Step 2: Evaluate the endpoints.

$$f(1)=1^2+1=2$$ $$f(2)=2^2+1=5$$

Step 3: Check the target value. The number \(3\) lies between \(2\) and \(5\).

Conclusion: By IVT, there exists some \(c\in[1,2]\) such that

$$f(c)=3.$$

In fact, if we solve it, we get

$$x^2+1=3 \Rightarrow x^2=2 \Rightarrow x=\sqrt{2}.$$

And \(\sqrt{2}\approx1.41\), which is indeed in \([1,2]\). But even without solving exactly, IVT already guaranteed that such a value existed.

Worked Example 3: Using IVT carefully with continuity

Can IVT be used for the function

$$f(x)=\frac{1}{x-1}$$

on the interval \([0,2]\) to show that \(f(x)=0\) for some \(x\)?

First, check continuity. The function \(\frac{1}{x-1}\) is not continuous on \([0,2]\) because it is undefined at \(x=1\).

Now check the endpoint values:

$$f(0)=\frac{1}{-1}=-1,? wait$$

Actually,

$$f(0)=-1, \qquad f(2)=1.$$

So \(0\) lies between \(-1\) and \(1\). However, we still cannot use IVT, because the function is not continuous on the full interval \([0,2]\).

And in fact, \(\frac{1}{x-1}=0\) has no solution, because a fraction with numerator \(1\) can never equal \(0\).

Lesson from this example: opposite signs at the endpoints are not enough by themselves. Continuity on the whole interval is required.

Worked Example 4: Writing a full IVT proof

Prove that the equation

$$2x^3-3x-1=0$$

has at least one solution in the interval \([1,2]\).

Solution:

Let

$$f(x)=2x^3-3x-1.$$

Since \(f\) is a polynomial, it is continuous on \([1,2]\).

Now evaluate the function at the endpoints:

$$f(1)=2(1)^3-3(1)-1=2-3-1=-2$$ $$f(2)=2(2)^3-3(2)-1=16-6-1=9$$

Because \(-2<0<9\), the number \(0\) lies between \(f(1)\) and \(f(2)\).

Therefore, by the Intermediate Value Theorem, there exists some \(c\in(1,2)\) such that

$$f(c)=0.$$

So the equation \(2x^3-3x-1=0\) has at least one real root in \((1,2)\).

This is a strong example of how to write a complete theorem-based argument: define the function, state continuity, evaluate the endpoints, and apply IVT.

Important notes and common mistakes

• IVT guarantees existence, not uniqueness. A function may cross the target value more than once.
• You must check continuity on the entire interval. Continuity only at the endpoints is not enough.
• IVT does not find the exact value of \(c\). It only proves that such a value exists.
• The target value can be any number \(N\), not just \(0\).

For example, if a continuous function has \(f(1)=4\) and \(f(3)=10\), then IVT guarantees that the function takes on every value between \(4\) and \(10\), such as \(5\), \(6.5\), or \(9.9\), somewhere between \(x=1\) and \(x=3\).

How IVT connects to limits and continuity

In the topic of limits and continuity, we study what happens to a function near points and across intervals. IVT is a major theorem that comes from continuity. If a function behaves continuously, then its outputs must also behave continuously, with no missing values between two outputs.

This idea is especially useful near boundaries of intervals, when checking whether a function must cross an axis or hit a certain height. Even if the exact value is hard to calculate, IVT gives a reliable existence result.

Quick checklist for IVT problems

• Is the function continuous on the interval?
• What are the values at the endpoints?
• Is the desired value between those endpoint outputs?
• If yes, state clearly that IVT guarantees some \(c\) in the interval.

Brief Summary

The Intermediate Value Theorem says that if a function is continuous on \([a,b]\), then it takes on every value between \(f(a)\) and \(f(b)\). A very common use is to prove a root exists: if \(f(a)\) and \(f(b)\) have opposite signs, then there is some \(c\in(a,b)\) with \(f(c)=0\). Always remember that continuity on the whole interval is the key condition.