# How to Evaluate Limits: 5 Methods With Worked Examples

> Evaluate limits step by step: direct substitution, factor and cancel, rationalizing, special trig limits, and a flowchart for picking a method.

Source: https://practicecalculus.com/blog/how-to-evaluate-limits
Updated: 2026-10-16

## Key takeaways

- Always try direct substitution first: if it gives a real number, that number is the limit.
- A $\frac{0}{0}$ result is not an answer. It tells you to simplify by factoring, rationalizing or using a known trig limit.
- A nonzero number over $0$ signals a vertical asymptote, so check the sign from each side.
- The two key trig limits are $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.

You know what a limit *means*, but when a homework problem says "evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$," what do you actually do? The good news is that almost every limit in a first calculus course yields to one of a handful of techniques, and there is a reliable way to pick the right one.

This guide gives you each method, a fully worked example for each, and a decision flowchart (in words) you can follow on any exam.

## The first rule: always try direct substitution

Most functions you meet (polynomials, rational functions, roots, trig, exponentials and logarithms) are **continuous** wherever they are defined. For a continuous function, the limit is simply the function value:

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

So the first move is always the same: plug in $a$ and see what happens. There are only three possible outcomes:

| Substitution gives | What it means | What to do |
| --- | --- | --- |
| A real number, like $7$ | The limit is that number | Stop. You are done. |
| $\frac{0}{0}$ | Indeterminate form | Simplify, then substitute again |
| $\frac{k}{0}$ with $k \neq 0$ | Vertical asymptote | Check signs from each side |

### Worked example 1: direct substitution

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

**Step 1. Substitute.** $2^2 + 3(2) - 1 = 4 + 6 - 1 = 9$.

**Step 2. Check the result.** It is a real number, so

$$\lim_{x \to 2} (x^2 + 3x - 1) = 9$$

The same works for $\lim_{x \to \pi} \cos x = \cos \pi = -1$ and $\lim_{x \to 1} \frac{x + 4}{x + 1} = \frac{5}{2}$. If substitution works, no other technique is needed.

## Method 2: factor and cancel

When substitution produces $\frac{0}{0}$, the numerator and denominator both vanish at $x = a$. For polynomials, that means both contain the factor $(x - a)$. Factor it out, cancel it, and substitute again.

Canceling is legal because a limit as $x \to a$ never uses $x = a$ itself, so $x - a \neq 0$ throughout the calculation.

### Worked example 2: a difference of squares

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

**Step 1. Substitute.** $\frac{9 - 9}{3 - 3} = \frac{0}{0}$. Indeterminate, so simplify.

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

**Step 3. Cancel the common factor.**

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

**Step 4. Substitute again.**

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

### Worked example 3: factoring a trinomial

Evaluate $\lim_{x \to -2} \frac{x^2 + 5x + 6}{x^2 - 4}$.

**Step 1. Substitute.** Top: $4 - 10 + 6 = 0$. Bottom: $4 - 4 = 0$. Indeterminate.

**Step 2. Factor both.** $x^2 + 5x + 6 = (x + 2)(x + 3)$ and $x^2 - 4 = (x + 2)(x - 2)$.

**Step 3. Cancel and substitute.**

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

> **Tip:** if substitution gives $\frac{0}{0}$ at $x = a$ for a polynomial ratio, you *know* $(x - a)$ is a factor of both top and bottom. Use that to guide your factoring (or polynomial division).

## Method 3: rationalize with the conjugate

When the $\frac{0}{0}$ comes from a square root in a sum or difference, factoring will not help. Instead multiply the numerator and denominator by the **conjugate**. The conjugate of $\sqrt{A} - B$ is $\sqrt{A} + B$, and their product is $A - B^2$, which has no root.

### Worked example 4: rationalizing the numerator

Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$.

**Step 1. Substitute.** $\frac{\sqrt{4} - 2}{0} = \frac{0}{0}$. Indeterminate.

**Step 2. Multiply by the conjugate over itself.**

$$\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \frac{(x + 4) - 4}{x\left(\sqrt{x+4} + 2\right)}$$

**Step 3. Simplify the top.** $(x+4) - 4 = x$, so the expression becomes

$$\frac{x}{x\left(\sqrt{x+4} + 2\right)} = \frac{1}{\sqrt{x+4} + 2}$$

**Step 4. Substitute again.**

$$\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$$

Notice that we deliberately did **not** multiply out the bottom. Leaving $x(\sqrt{x+4}+2)$ factored is what made the cancellation visible.

## Method 4: special trig limits

Trig expressions often give $\frac{0}{0}$ in a way no algebra can untangle. For these you rely on two standard limits, both with $x$ in radians:

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

The first is usually proved with the squeeze theorem and a picture of the unit circle. The graph shows what it says: $\frac{\sin x}{x}$ is undefined at $0$, but it heads straight to $1$.

```graph
title: 'Graph of y = sin(x)/x with a hole at (0, 1)'
caption: 'The function is undefined at $x = 0$, yet both sides approach $1$, so $\lim_{x \to 0} \frac{\sin x}{x} = 1$.'
x: [-12, 12]
y: [-0.5, 1.4]
functions:
  - expr: sin(x)/x
    label: y = sin(x)/x
points:
  - [0, 1, '(0, 1)', open]
```

The trick with these limits is to make the argument of the sine match the denominator exactly.

### Worked example 5: adjusting the argument

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

**Step 1. Substitute.** $\frac{\sin 0}{0} = \frac{0}{0}$.

**Step 2. Make the denominator match the argument.** Multiply top and bottom by $5$:

$$\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}$$

**Step 3. Use the special limit.** As $x \to 0$, $u = 5x \to 0$ too, so $\frac{\sin(5x)}{5x} \to 1$:

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

The same idea gives $\lim_{x \to 0} \frac{\sin(3x)}{\sin(7x)} = \frac{3}{7}$ (write it as $\frac{3}{7} \cdot \frac{\sin 3x}{3x} \cdot \frac{7x}{\sin 7x}$) and $\lim_{x \to 0} \frac{\tan x}{x} = 1$ (write $\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$, which approaches $1 \cdot 1$).

## Method 5: nonzero over zero means an asymptote

If substitution gives a **nonzero** number over $0$, no amount of algebra will produce a finite limit. The function blows up. Your job is to find the direction on each side.

### Worked example 6: checking signs

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

**Step 1. Substitute.** $\frac{3}{0}$. Nonzero over zero, so there is a vertical asymptote at $x = 2$.

**Step 2. Right side.** For $x$ slightly more than $2$ (say $2.01$), the top is about $3$ and the bottom is a tiny positive number, so the fraction is a huge positive number: $\lim_{x \to 2^+} = +\infty$.

**Step 3. Left side.** For $x$ slightly less than $2$, the bottom is a tiny negative number, so $\lim_{x \to 2^-} = -\infty$.

**Step 4. Conclude.** The sides disagree, so $\lim_{x \to 2} \frac{x + 1}{x - 2}$ does not exist. (If both sides had gone to $+\infty$, as with $\frac{1}{(x-2)^2}$, you would write $+\infty$.)

## A decision flowchart for any limit

Here is the whole process as a sequence of questions. Run through it in order.

1. **Is $x$ heading to $\pm\infty$?** If yes, this is a limit at infinity: compare the highest powers (see [limits at infinity](/blog/limits-at-infinity)).
2. **Is the function piecewise, or does it contain an absolute value?** If yes, compute the left and right limits separately and compare.
3. **Substitute.** If you get a real number, that is the answer. Stop.
4. **Got $\frac{k}{0}$ with $k \neq 0$?** Vertical asymptote. Check the sign from each side.
5. **Got $\frac{0}{0}$?** Look at what the expression contains:
   - Polynomials: **factor and cancel**.
   - A square root in a sum or difference: **multiply by the conjugate**.
   - Fractions inside fractions: **combine over a common denominator**, then cancel.
   - $\sin$, $\cos$ or $\tan$ of something going to $0$: **use the special trig limits**.
6. **Still stuck, and you know derivatives?** Use [L'Hôpital's rule](/blog/lhopitals-rule).
7. After simplifying, **go back to step 3** and substitute again.

## Common mistakes when evaluating limits

1. **Writing "$= \frac{0}{0}$" as the answer.** $\frac{0}{0}$ means "keep going," never "the limit is undefined."
2. **Dropping the $\lim$ symbol too early.** Write $\lim_{x \to a}$ on every line until you actually substitute. Many teachers deduct points for $\frac{x^2-9}{x-3} = x + 3 = 6$, which is false as written.
3. **Canceling terms instead of factors.** In $\frac{x^2 - 9}{x - 3}$ you cannot cancel the $x$'s or the $9$ with the $3$. Only whole factors cancel.
4. **Using degrees.** $\lim_{x \to 0} \frac{\sin x}{x} = 1$ is true only in radians.
5. **Forgetting to check both sides at an asymptote.** $\frac{k}{0}$ alone does not tell you the sign of the infinity.

## Quick reference

| Situation | Technique | Example | Limit |
| --- | --- | --- | --- |
| Substitution works | Direct substitution | $\lim_{x \to 2} (x^2 + 3x - 1)$ | $9$ |
| $\frac{0}{0}$, polynomials | Factor and cancel | $\lim_{x \to 3} \frac{x^2-9}{x-3}$ | $6$ |
| $\frac{0}{0}$, square root | Conjugate | $\lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}$ | $\frac{1}{4}$ |
| $\frac{0}{0}$, trig | Special limits | $\lim_{x \to 0} \frac{\sin 5x}{x}$ | $5$ |
| $\frac{k}{0}$ | Sign check | $\lim_{x \to 2} \frac{x+1}{x-2}$ | does not exist |

## Make these methods automatic

Recognizing which technique a limit needs is a pattern-matching skill, and patterns come from repetition. The interactive lesson below throws a steady mix of factor-and-cancel problems at you, checks every step instantly, and shows the full solution whenever you get stuck. Ten minutes of practice will make the flowchart above second nature.

## Frequently asked questions

### What is the first step in evaluating a limit?

Plug the value into the function (direct substitution). If you get an ordinary real number, you are done, because polynomials, rational functions, roots, trig, exponential and log functions are continuous wherever they are defined.

### What does it mean when a limit gives 0/0?

$\frac{0}{0}$ is an indeterminate form: it tells you nothing yet about the limit. It usually means a common factor is hiding in the numerator and denominator. Factor and cancel, rationalize, or use a special limit, then substitute again.

### What do you do when a limit gives a number divided by zero?

If substitution gives something like $\frac{5}{0}$, the function has a vertical asymptote there. Check the sign of the expression just to the left and just to the right. If both sides go to the same infinity you can write $\infty$ or $-\infty$; if they differ, the two-sided limit does not exist.

### When should you rationalize a limit?

Rationalize when a square root appears in a sum or difference, like $\sqrt{x+4} - 2$, and substitution gives $\frac{0}{0}$. Multiply the top and bottom by the conjugate, $\sqrt{x+4} + 2$, to clear the root from the troublesome part.

### Can I use L'Hôpital's rule instead of algebra?

For $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms you can, once you know derivatives. But the algebraic methods are usually faster for simple rational functions, and many courses expect them before derivatives are introduced.