Lesson: The Operations of Limits

1. What Is a Limit?

In calculus, a limit describes what value a function gets close to as the input gets close to a certain number.

The notation looks like this:

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

This is read as:

“The limit of $f(x)$ as $x$ approaches $a$ is $L$.”

This means that as $x$ gets closer and closer to $a$, the output $f(x)$ gets closer and closer to $L$.

A limit is not always asking what happens exactly at $x=a$. Instead, it asks what happens near $x=a$.

For example:

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

As $x$ gets closer to $2$, the expression $x+3$ gets closer to:

$$ 2+3=5 $$

So:

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

2. Direct Substitution

The easiest way to evaluate many limits is by using direct substitution.

Direct substitution means replacing $x$ with the number that $x$ is approaching.

Example 1

Evaluate:

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

Substitute $x=4$:

$$ 2(4)+1=8+1=9 $$

Therefore:

$$ \lim_{x \to 4} (2x+1)=9 $$

Example 2

Evaluate:

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

Substitute $x=-3$:

$$ (-3)^2+5(-3)=9-15=-6 $$

Therefore:

$$ \lim_{x \to -3} (x^2+5x)=-6 $$

Direct substitution works very well for simple functions like polynomials.

A polynomial is an expression such as:

$$ x^2+3x-7 $$

or

$$ 4x^3-2x+9 $$

For polynomial functions, you can usually plug in the value directly.

3. Limit Operation Rules

Limits have operation rules that make them easier to work with.

Suppose:

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

and

$$ \lim_{x \to a} g(x)=B $$

Then the following rules are true.

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Addition Rule

The limit of a sum is the sum of the limits:

$$ \lim_{x \to a} [f(x)+g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) $$

So:

$$ \lim_{x \to a} [f(x)+g(x)] = A+B $$

Example

Evaluate:

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

Break it into two limits:

$$ \lim_{x \to 2} x^2 + \lim_{x \to 2} 3x $$

Now substitute:

$$ 2^2 + 3(2) $$

$$ 4+6=10 $$

Therefore:

$$ \lim_{x \to 2} (x^2+3x)=10 $$

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Subtraction Rule

The limit of a difference is the difference of the limits:

$$ \lim_{x \to a} [f(x)-g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) $$

So:

$$ \lim_{x \to a} [f(x)-g(x)] = A-B $$

Example

Evaluate:

$$ \lim_{x \to 5} (x^2-4x) $$

Substitute $x=5$:

$$ 5^2-4(5) $$

$$ 25-20=5 $$

Therefore:

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

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Multiplication Rule

The limit of a product is the product of the limits:

$$ \lim_{x \to a} [f(x)g(x)] = \left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} g(x)\right) $$

So:

$$ \lim_{x \to a} [f(x)g(x)] = AB $$

Example

Evaluate:

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

Substitute $x=3$:

$$ (3)(3+2) $$

$$ 3(5)=15 $$

Therefore:

$$ \lim_{x \to 3} x(x+2)=15 $$

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Constant Multiple Rule

If a function is multiplied by a constant, the constant can stay outside the limit:

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

Here, $c$ is a constant number.

Example

Evaluate:

$$ \lim_{x \to 4} 7x^2 $$

The constant is $7$. Substitute $x=4$:

$$ 7(4^2) $$

$$ 7(16)=112 $$

Therefore:

$$ \lim_{x \to 4} 7x^2=112 $$

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Division Rule

The limit of a quotient is the quotient of the limits:

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

So:

$$ \lim_{x \to a} \frac{f(x)}{g(x)}=\frac{A}{B} $$

This rule only works if:

$$ B \neq 0 $$

That means the limit of the denominator cannot be zero.

Example

Evaluate:

$$ \lim_{x \to 2} \frac{x+6}{x+1} $$

Substitute $x=2$:

$$ \frac{2+6}{2+1} $$

$$ \frac{8}{3} $$

Therefore:

$$ \lim_{x \to 2} \frac{x+6}{x+1}=\frac{8}{3} $$

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Power Rule

If a function is raised to a power, the limit can also be raised to that power:

$$ \lim_{x \to a} [f(x)]^n = \left(\lim_{x \to a} f(x)\right)^n $$

Example

Evaluate:

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

Substitute $x=2$:

$$ (2+1)^3 $$

$$ 3^3=27 $$

Therefore:

$$ \lim_{x \to 2} (x+1)^3=27 $$

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Root Rule

For roots, we can usually substitute directly as long as the root is defined.

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

Example

Evaluate:

$$ \lim_{x \to 9} \sqrt{x} $$

Substitute $x=9$:

$$ \sqrt{9}=3 $$

Therefore:

$$ \lim_{x \to 9} \sqrt{x}=3 $$

Another example:

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

Substitute $x=4$:

$$ \sqrt{4+5}=\sqrt{9}=3 $$

Therefore:

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

4. What Happens When Direct Substitution Gives $\frac{0}{0}$?

Sometimes direct substitution does not immediately work.

For example:

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

If we substitute $x=3$, we get:

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

$$ \frac{9-9}{0} $$

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

The expression $\frac{0}{0}$ is called an indeterminate form.

This does not mean the limit does not exist. It means we need to simplify first.

Notice that the numerator can be factored:

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

So:

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

Now we can cancel the common factor:

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

So the original limit becomes:

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

Now substitute:

$$ 3+3=6 $$

Therefore:

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

The important idea is that when direct substitution gives $\frac{0}{0}$, try simplifying the expression.

5. Factoring to Evaluate Limits

Factoring is one of the most common techniques for evaluating limits.

Example

Evaluate:

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

First, try direct substitution:

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

Since we got $\frac{0}{0}$, we factor the numerator:

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

Now rewrite the expression:

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

Cancel the common factor:

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

Now evaluate:

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

Therefore:

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

Even though the original function was undefined at $x=2$, the limit still exists because the function approaches $4$ near $x=2$.

6. Limits Involving Fractions

When evaluating a limit with a fraction, always check the denominator.

If the denominator does not become zero, direct substitution usually works.

Example

Evaluate:

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

Substitute $x=1$:

$$ \frac{1^2+3}{1+4} $$

$$ \frac{1+3}{5} $$

$$ \frac{4}{5} $$

Therefore:

$$ \lim_{x \to 1} \frac{x^2+3}{x+4}=\frac{4}{5} $$

But if the denominator becomes zero, we need to be more careful.

For example:

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

Substitute $x=2$:

$$ \frac{2+1}{2-2} = \frac{3}{0} $$

This is not the same as $\frac{0}{0}$. Since the numerator is not also zero, this often means the limit may not exist or may become very large in the positive or negative direction.

At this beginner stage, remember:

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

usually means “try simplifying,” while

$$ \frac{\text{nonzero}}{0} $$

means “be careful; the limit may not exist.”

7. One-Sided Limits

Sometimes we only care about what happens from one side.

A left-hand limit looks like this:

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

This means $x$ approaches $a$ from the left side, using values less than $a$.

A right-hand limit looks like this:

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

This means $x$ approaches $a$ from the right side, using values greater than $a$.

For the two-sided limit to exist, the left-hand limit and right-hand limit must be equal.

In other words:

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

exists only if:

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

Example

Suppose:

As $x$ approaches $1$ from the left, $f(x)$ approaches $2$:

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

As $x$ approaches $1$ from the right, $f(x)$ approaches $5$:

$$ \lim_{x \to 1^+} f(x)=5 $$

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

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

8. Summary of Limit Operation Rules

Here are the main limit rules:

$$ \lim_{x \to a} [f(x)+g(x)] = \lim_{x \to a} f(x)+\lim_{x \to a} g(x) $$

$$ \lim_{x \to a} [f(x)-g(x)] = \lim_{x \to a} f(x)-\lim_{x \to a} g(x) $$

$$ \lim_{x \to a} [f(x)g(x)] = \left(\lim_{x \to a} f(x)\right) \left(\lim_{x \to a} g(x)\right) $$

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

as long as:

$$ \lim_{x \to a} g(x) \neq 0 $$

$$ \lim_{x \to a} [f(x)]^n = \left(\lim_{x \to a} f(x)\right)^n $$

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

when the square root is defined.

9. Step-by-Step Strategy for Evaluating Limits

When evaluating a limit, use this process:

• Try direct substitution.
• If direct substitution gives a real number, that is usually the answer.
• If direct substitution gives $\frac{0}{0}$, simplify the expression.
• Try factoring, expanding, or canceling common factors.
• After simplifying, substitute again.

If one-sided limits are involved, check the left side and right side separately.

10. Practice Examples

Practice Example 1

Evaluate:

$$ \lim_{x \to 6} (x^2-2x+1) $$

Substitute $x=6$:

$$ 6^2-2(6)+1 $$

$$ 36-12+1=25 $$

Therefore:

$$ \lim_{x \to 6} (x^2-2x+1)=25 $$

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Practice Example 2

Evaluate:

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

Substitute $x=3$:

$$ \frac{3+4}{3^2+1} $$

$$ \frac{7}{9+1} $$

$$ \frac{7}{10} $$

Therefore:

$$ \lim_{x \to 3} \frac{x+4}{x^2+1}=\frac{7}{10} $$

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Practice Example 3

Evaluate:

$$ \lim_{x \to 5} \frac{x^2-25}{x-5} $$

First substitute:

$$ \frac{5^2-25}{5-5} = \frac{25-25}{0} = \frac{0}{0} $$

Since this gives $\frac{0}{0}$, factor the numerator:

$$ x^2-25=(x-5)(x+5) $$

Rewrite:

$$ \frac{x^2-25}{x-5} = \frac{(x-5)(x+5)}{x-5} $$

Cancel:

$$ x+5 $$

Now substitute:

$$ 5+5=10 $$

Therefore:

$$ \lim_{x \to 5} \frac{x^2-25}{x-5}=10 $$

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Practice Example 4

Evaluate:

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

Substitute $x=1$:

$$ \frac{1^2+2(1)+1}{1+1} $$

$$ \frac{1+2+1}{2} $$

$$ \frac{4}{2}=2 $$

Therefore:

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

11. Final Key Idea

Limits are about what a function approaches, not always what the function equals.

The main question is:

“As $x$ gets closer and closer to a certain number, what does $f(x)$ get closer and closer to?”

Most beginner limit problems can be solved by direct substitution. If direct substitution gives $\frac{0}{0}$, simplify the expression and try again.