How to Find Limits (With Examples)

Limits describe what a function approaches, not necessarily what it equals. This guide walks through the most common limit techniques with clear steps and many examples. If you are new to limits, start with direct substitution and then move into indeterminate forms like $0/0$.

Quick Roadmap

1. Try direct substitution.
2. If you see an indeterminate form ($0/0$ or $\infty/\infty$), simplify:
   • Factor and cancel
   • Rationalize
   • Use trig identities
   • Combine into a single fraction
3. For limits at infinity, compare dominant terms.
4. For piecewise functions, check left and right limits.

What a Limit Means

When we write

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

we mean that the values of $f(x)$ get closer and closer to $L$ as $x$ gets closer and closer to $a$. The function value at $a$ might be different or even undefined, and the limit can still exist.

1. Direct Substitution (Start Here)

If $f(x)$ is continuous at $a$, just plug in $x = a$.

Example 1

$$\lim_{x \to 3} (x^2 - 4x + 7)$$

Solution

$$3^2 - 4(3) + 7 = 9 - 12 + 7 = 4$$

Answer: $4$

Example 2

$$\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 3}$$

Solution

Direct substitution works because the denominator is not zero:

$$\frac{(-2)^2 + 5(-2) + 6}{-2 + 3} = \frac{4 - 10 + 6}{1} = 0$$

Answer: $0$

2. Factoring and Canceling ($0/0$ Forms)

When substitution gives $0/0$, it means there is a removable factor.

Example 3

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

Solution

Factor:

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

Cancel $x - 2$:

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

Answer: $4$

Example 4

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

Solution

Factor sum of cubes:

$$x^3 + 1 = (x + 1)(x^2 - x + 1)$$

Cancel $x + 1$:

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

Answer: $3$

3. Rationalizing (Roots in the Numerator or Denominator)

If you have a square root and get $0/0$, multiply by the conjugate.

Example 5

$$\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$$

Solution

Multiply by the conjugate:

$$\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}$$

Simplify:

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

Cancel $x$:

$$\lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{6}$$

Answer: $\frac{1}{6}$

Example 6

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

Solution

Rewrite and rationalize:

$$\frac{4 - \sqrt{x}}{4 - x} \cdot \frac{4 + \sqrt{x}}{4 + \sqrt{x}} = \frac{16 - x}{(4 - x)(4 + \sqrt{x})}$$

Now $16 - x = -(x - 16)$ and $4 - x = -(x - 4)$, so:

$$\frac{16 - x}{(4 - x)(4 + \sqrt{x})} = \frac{1}{4 + \sqrt{x}}$$

Plug in $x = 4$:

$$\frac{1}{4 + 2} = \frac{1}{6}$$

Answer: $\frac{1}{6}$

4. Trigonometric Limits and Identities

Two key limits:

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

Also:

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

Example 7

$$\lim_{x \to 0} \frac{\sin(5x)}{x}$$

Solution

Multiply and divide by $5$:

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

Answer: $5$

Example 8

$$\lim_{x \to 0} \frac{1 - \cos(3x)}{x^2}$$

Solution

Use the known limit:

$$\frac{1 - \cos(3x)}{x^2} = \frac{1 - \cos(3x)}{(3x)^2} \cdot 9$$

So:

$$9 \cdot \lim_{x \to 0} \frac{1 - \cos(3x)}{(3x)^2} = 9 \cdot \frac{1}{2} = \frac{9}{2}$$

Answer: $\frac{9}{2}$

5. Combine and Simplify Complex Fractions

Sometimes you need to combine terms first.

Example 9

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

Solution

Combine the numerator:

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

Note $2 - x = -(x - 2)$:

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

Now substitute:

$$-\frac{1}{2(2)} = -\frac{1}{4}$$

Answer: $-\frac{1}{4}$

6. Limits at Infinity

For rational functions, compare the highest power of $x$.

Example 10

$$\lim_{x \to \infty} \frac{4x^3 - 2x + 1}{2x^3 + 7x^2 - 5}$$

Solution

Divide by $x^3$:

$$\lim_{x \to \infty} \frac{4 - \frac{2}{x^2} + \frac{1}{x^3}}{2 + \frac{7}{x} - \frac{5}{x^3}} = \frac{4}{2} = 2$$

Answer: $2$

Example 11

$$\lim_{x \to \infty} \frac{3x^2 + 1}{5x^3 - 2x}$$

Solution

Degree in denominator is higher, so limit is $0$.

Answer: $0$

7. One-Sided Limits and Piecewise Functions

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

Example 12

Let

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

Find $\lim_{x \to 1} f(x)$.

Solution

Left-hand limit:

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

Right-hand limit:

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

They are not equal, so the limit does not exist.

Answer: DNE

8. Squeeze Theorem (Bounds Method)

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

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

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

Example 13

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

Solution

Since $-1 \le \sin(1/x) \le 1$,

$$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$$

Both bounds go to $0$, so the limit is $0$.

Answer: $0$

9. A Practical Checklist for Any Limit

Use this checklist when you get stuck:

1. Substitute directly and see what happens.
2. If $0/0$, try factoring or common denominators.
3. If roots are involved, rationalize.
4. If trig is involved, use special trig limits or identities.
5. If infinity appears, compare highest powers or divide by the highest power.
6. For piecewise functions, evaluate left and right limits.
7. Consider Squeeze Theorem if the function oscillates.

Common Mistakes to Avoid

• Cancelling terms that are not factors.
• Plugging in too early when you have $0/0$.
• Forgetting to check one-sided limits for piecewise or absolute value functions.
• Mixing degrees incorrectly in limits at infinity.
• Treating $\lim$ like direct evaluation when the function is undefined at the point.

Practice Problems (With Answers)

1. $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
   Answer: $2$
2. $\lim_{x \to 0} \frac{\sin(2x)}{x}$
   Answer: $2$
3. $\lim_{x \to 0} \frac{\sqrt{4 + x} - 2}{x}$
   Answer: $\frac{1}{4}$
4. $\lim_{x \to \infty} \frac{7x^2 - 3}{2x^2 + 5x}$
   Answer: $\frac{7}{2}$
5. $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$
   Answer: $\frac{1}{2}$

Final Takeaway

Limits become much easier when you follow a consistent process. Start with direct substitution, then resolve indeterminate forms using algebra or identities. With practice, you will recognize patterns quickly and choose the right technique on the first try.