How to Find Limits: 5 Methods With Examples

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

This reads "the limit of $f(x)$ as $x$ approaches $a$ is $L$." It means that as $x$ gets close to $a$ (from both sides), $f(x)$ gets close to $L$.

The key point: a limit cares about what happens near $a$, not at $a$. The function does not even have to be defined at $a$. For example, $f(x) = \frac{x^2 - 9}{x - 3}$ is undefined at $x = 3$, but its limit there is 6, as you will see below.

Always start by plugging the value in. For polynomials, roots, exponentials, and trig functions where the function is defined, this is all you need.

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

1. Substitute $x = 2$: $2^2 + 3(2) - 1 = 4 + 6 - 1$.
2. Answer: $9$.

What the result of substitution tells you:

• A real number, like 9: that is the limit. You are done.
• $\frac{0}{0}$: this is an indeterminate form. The limit could be anything, so use factoring, rationalizing, or L'Hôpital's rule.
• A nonzero number over 0, like $\frac{5}{0}$: the function shoots off toward $\infty$ or $-\infty$. Check the one-sided limits (see the last section).

If substitution gives $\frac{0}{0}$ and you see polynomials, factor the top and bottom and cancel the common factor.

Example 1: Find $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

1. Substitute: $\frac{9 - 9}{3 - 3} = \frac{0}{0}$. Indeterminate, so keep going.
2. Factor the top as a difference of squares: $\frac{(x - 3)(x + 3)}{x - 3}$.
3. Cancel $(x - 3)$. This is allowed because $x \neq 3$ while $x$ approaches 3.
4. Substitute into what is left: $\lim_{x \to 3} (x + 3) = 6$.

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

1. Substitute: $\frac{4 - 10 + 6}{4 - 4} = \frac{0}{0}$.
2. Factor: $\frac{(x + 2)(x + 3)}{(x + 2)(x - 2)}$.
3. Cancel $(x + 2)$: $\frac{x + 3}{x - 2}$.
4. Substitute: $\frac{-2 + 3}{-2 - 2} = \frac{1}{-4} = -\frac{1}{4}$.

If substitution gives $\frac{0}{0}$ and there is a square root, multiply the top and bottom by the conjugate. The conjugate of $\sqrt{A} - B$ is $\sqrt{A} + B$.

Example: Find $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.

1. Substitute: $\frac{\sqrt{9} - 3}{0} = \frac{0}{0}$.
2. Multiply by the conjugate over itself:

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

3. The top becomes a difference of squares: $(x + 9) - 9 = x$.
4. So the expression is $\frac{x}{x(\sqrt{x + 9} + 3)}$. Cancel $x$: $\frac{1}{\sqrt{x + 9} + 3}$.
5. Substitute: $\frac{1}{\sqrt{9} + 3} = \frac{1}{6}$.

Do not expand the bottom. Leaving it factored is what lets the $x$ cancel.

For a rational function as $x \to \infty$, divide every term by the highest power of $x$ in the denominator. Any term like $\frac{c}{x}$ or $\frac{c}{x^2}$ goes to 0.

Example: Find $\lim_{x \to \infty} \frac{3x^2 - x + 2}{5x^2 + 4}$.

1. The highest power in the denominator is $x^2$. Divide every term by $x^2$:

$$\frac{3 - \frac{1}{x} + \frac{2}{x^2}}{5 + \frac{4}{x^2}}$$

2. As $x \to \infty$, the fractions with $x$ in the bottom go to 0.
3. Answer: $\frac{3}{5}$.

Degree shortcut (compare the highest powers on top and bottom):

• Top degree smaller: the limit is 0. Example: $\lim_{x \to \infty} \frac{2x + 1}{x^2 - 3} = 0$.
• Degrees equal: the limit is the ratio of the leading coefficients, like $\frac{3}{5}$ above.
• Top degree bigger: there is no finite limit. The function grows toward $\infty$ or $-\infty$.

If substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, you can differentiate the top and bottom separately and try again:

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

Example 1: Find $\lim_{x \to 0} \frac{\sin x}{x}$.

1. Substitute: $\frac{0}{0}$. The rule applies.
2. Differentiate top and bottom: $\frac{\cos x}{1}$.
3. Substitute: $\cos 0 = 1$.

Example 2: Find $\lim_{x \to 0} \frac{e^x - 1}{x}$.

1. Substitute: $\frac{1 - 1}{0} = \frac{0}{0}$.
2. Differentiate: $\frac{e^x}{1}$.
3. Substitute: $e^0 = 1$.

This is not the quotient rule. You take the derivative of the numerator and the derivative of the denominator on their own. And you can only use it on $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

A two-sided limit exists only if the left-hand limit ($x \to a^-$) and the right-hand limit ($x \to a^+$) are the same number.

Example 1: $\lim_{x \to 0} \frac{1}{x}$. From the right, $\frac{1}{x}$ grows to $\infty$. From the left, it drops to $-\infty$. The sides disagree, so the limit does not exist.

Example 2: $\lim_{x \to 0} \frac{|x|}{x}$. For $x > 0$ the expression equals 1, and for $x < 0$ it equals $-1$. The one-sided limits are 1 and $-1$, so the limit does not exist.

A checklist for any limit:

1. Substitute first.
2. Real number? Done.
3. $\frac{0}{0}$ with polynomials? Factor and cancel.
4. $\frac{0}{0}$ with a square root? Multiply by the conjugate.
5. $x \to \infty$? Divide by the highest power in the denominator.
6. Still stuck on $\frac{0}{0}$ or $\frac{\infty}{\infty}$? Use L'Hôpital's rule.
7. Nonzero over 0? Check both one-sided limits.

Paste a limit into the VeilGPT solver to check your steps against your own work.