How to Evaluate Limits

To evaluate $lim_{x \to a} f (x)$ , first try substituting $x = a$ directly. If you get an indeterminate form, apply one of the four algebraic techniques described in this section.

Technique	When to use
Direct substitution	Always try first; works when $f$ is continuous at $a$
Factoring	$0 / 0$ form with polynomial numerator and denominator
Rationalizing	$0 / 0$ form involving a square root that vanishes
Common denominator	$\infty - \infty$ form involving fractions
Leading term	$\pm \infty / \pm \infty$ or $\pm \infty$ for polynomials/rational functions

Step 1: Try Direct Substitution

To evaluate $lim_{x \to a} f (x)$ , first try putting $x = a$ into $f (x)$ . If $f$ is continuous at $a$ , then $lim_{x \to a} f (x) = f (a)$ and you are done. Be careful with piecewise-defined functions, where direct substitution may give the wrong answer at a break point.

If substitution gives one of the indeterminate forms

\frac{0}{0}, \frac{\pm \infty}{\pm \infty}, 0 \cdot (\pm \infty), \infty - \infty,

then proceed to one of the techniques below. (The most powerful technique, L'Hôpital's Rule, is covered in the Applications of Differentiation chapter.)

Technique 1: Factoring

Factoring technique: If $P (x)$ and $Q (x)$ are polynomials and $P (a) = Q (a) = 0$ , then $(x - a)$ is a common factor of both. Factor out $(x - a)$ , cancel it from numerator and denominator, then substitute.

If factoring $(x - a)$ directly is difficult, you can divide both $P (x)$ and $Q (x)$ by $(x - a)$ using polynomial long division.

Find $lim_{x \to 2} \frac{x^{2} - 4}{3 x - 6}$ .

Solution

Substituting

x = 2

gives

0 / 0

. Since both

x^{2} - 4

and

3 x - 6

vanish at

x = 2

, factor out

(x - 2)

x^{2} - 4 = (x - 2) (x + 2), 3 x - 6 = 3 (x - 2) .

Therefore: \lim_{x \to 2} \frac{x^2-4}{3x-6} = \lim_{x \to 2} \frac{\cancel{(x-2)}(x+2)}{3\cancel{(x-2)}} = \frac{1}{3}\lim_{x \to 2}(x+2) = \frac{1}{3}(4) = \frac{4}{3}.

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

Solution

Substituting

x = 1

gives

0 / 0

. Use the Difference of Cubes formula

A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})

x^{3} - 1 = (x - 1) (x^{2} + x + 1) .

For the denominator, since

x = 1

is a root of

2 x^{2} + x - 3

(x - 1)

is a factor. Comparing

2 x^{2} + x - 3 = (x - 1) (a x + b) = a x^{2} + (b - a) x - b

, we get

a = 2, b = 3

2 x^{2} + x - 3 = (x - 1) (2 x + 3) .

Therefore: \frac{x^3-1}{2x^2+x-3} = \frac{\cancel{(x-1)}(x^2+x+1)}{\cancel{(x-1)}(2x+3)} = \frac{x^2+x+1}{2x+3} \quad (x \neq 1). Now substitute

x = 1

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

Technique 2: Rationalizing

Rationalizing technique: If the limit has the indeterminate form $0 / 0$ and the numerator or denominator contains an expression of the form $A - B \sqrt{C}$ that vanishes upon substitution, multiply both numerator and denominator by the conjugate $A + B \sqrt{C}$ and use $(A - B \sqrt{C}) (A + B \sqrt{C}) = A^{2} - B^{2} C$ .

Find $lim_{x \to - 4} \frac{x^{2} - 16}{3 - \sqrt{x^{2} - 7}}$ .

Solution

Substituting

x = - 4

gives

0 / 0

. Multiply numerator and denominator by the conjugate

3 + \sqrt{x^{2} - 7}

: \begin{aligned} \lim_{x \to -4} \frac{x^2-16}{3-\sqrt{x^2-7}} &= \lim_{x \to -4} \frac{(x^2-16)(3+\sqrt{x^2-7})}{3^2-(x^2-7)} \\ &= \lim_{x \to -4} \frac{(x^2-16)(3+\sqrt{x^2-7})}{16-x^2} \\ &= \lim_{x \to -4} \bigl(-3-\sqrt{x^2-7}\bigr) \\ &= -3 - \sqrt{16-7} = -3 - 3 = -6. \end{aligned}

Technique 3: Common Denominator

Common denominator technique: If the limit of $f \pm g$ has the indeterminate form $\infty - \infty$ and $f$ , $g$ are fractions, find a common denominator, combine into one fraction, and simplify.

Find $lim_{x \to 1} (\frac{2}{x^{2} - 1} - \frac{1}{x - 1})$ .

Solution

x \to 1^{+}

, both

\frac{2}{x^{2} - 1}

and

\frac{1}{x - 1}

approach

+ \infty

, giving

\infty - \infty

. As

x \to 1^{-}

, both approach

- \infty

, giving

- \infty - (- \infty)

. Use the common denominator

x^{2} - 1 = (x - 1) (x + 1)

: \begin{aligned} \lim_{x \to 1}\!\left(\frac{2}{x^2-1} - \frac{1}{x-1}\right) &= \lim_{x \to 1} \frac{2 - (x+1)}{(x-1)(x+1)} = \lim_{x \to 1} \frac{1-x}{(x-1)(x+1)} \\ &= \lim_{x \to 1} \frac{-\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} = \lim_{x \to 1} \frac{-1}{x+1} = -\frac{1}{2}. \end{aligned}

Technique 4: Leading Term (for $x \to \pm \infty$ )

Leading term technique: When $x$ is numerically large ( $x \to + \infty$ or $x \to - \infty$ ), the behavior of a polynomial or rational function is determined entirely by its leading term.

For polynomials ( $a_{n} \neq 0$ ):

lim_{x \to \pm \infty} (a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}) = lim_{x \to \pm \infty} a_{n} x^{n} .

For rational functions ( $a_{n} \neq 0$ , $b_{m} \neq 0$ ):

lim_{x \to \pm \infty} \frac{a_{n} x^{n} + \dots + a_{0}}{b_{m} x^{m} + \dots + b_{0}} = lim_{x \to \pm \infty} \frac{a_{n} x^{n}}{b_{m} x^{m}} .

Find $lim_{x \to - \infty} (- 3 x^{5} - 8 x^{2} + 37)$ .

Solution

Factor out the highest power

x^{5}

: \begin{aligned} \lim_{x \to -\infty}(-3x^5 - 8x^2 + 37) &= \lim_{x \to -\infty} x^5\!\left(-3 - \frac{8}{x^3} + \frac{37}{x^5}\right) \\ &= \left(\lim_{x \to -\infty} x^5\right)\!\left(-3 - 0 + 0\right) \\ &= (-\infty)(-3) = +\infty. \end{aligned}

Find:

$lim_{x \to - \infty} \frac{3 x^{3} - 4 x^{2} + x + 3}{- x^{3} + 2 x^{2}}$
$lim_{x \to + \infty} \frac{13 x^{4} - 5 x^{3} + 4 x + 15}{- x^{5} + 16 x^{2} + 1}$

Solution

**(a)** Factor out

x^{3}

from numerator and denominator:

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

Using the leading-term shortcut directly:

lim_{x \to - \infty} \frac{3 x^{3}}{- x^{3}} = - 3

. **(b)** Keep only leading terms:

lim_{x \to + \infty} \frac{13 x^{4}}{- x^{5}} = lim_{x \to + \infty} \frac{- 13}{x} = 0.

Frequently Asked Questions

What do I do if none of the four techniques work?

If factoring, rationalizing, common denominator, and leading-term techniques all fail to resolve an indeterminate form, the most powerful remaining tool is **L'Hôpital's Rule**, which handles

0 / 0

and

\pm \infty / \pm \infty

by differentiating numerator and denominator separately. This is covered in the Applications of Differentiation chapter.

When I factor and cancel

(x - a)

, am I changing the function?

Strictly speaking, yes: the simplified expression differs from the original at

x = a

(the original is undefined there; the simplified one may be defined). However, the limit as

x \to a

is unchanged, because limits ignore what happens at

x = a

exactly. So the cancellation is valid for the purpose of evaluating the limit.

How do I evaluate a limit involving

\sqrt{x}

x \to \infty

Factor out the appropriate power of

x

from under the radical. For example, to find

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

, factor

x^{4}

from inside the square root:

\sqrt{4 x^{4} + 3 x^{2} + 1} = x^{2} \sqrt{4 + 3 / x^{2} + 1 / x^{4}}

(note:

\sqrt{x^{4}} = x^{2}

since

x^{2} \geq 0

). Then simplify.

Step 1: Try Direct Substitution

Technique 1: Factoring

Technique 2: Rationalizing

Technique 3: Common Denominator

Technique 4: Leading Term (for $x \to \pm \infty$ )

Frequently Asked Questions

Quick Links

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How to Evaluate Limits

Step 1: Try Direct Substitution

Technique 1: Factoring

Technique 2: Rationalizing

Technique 3: Common Denominator

Technique 4: Leading Term (for x → ± ∞ )

Frequently Asked Questions

Quick Links

Social Media

Technique 4: Leading Term (for $x \to \pm \infty$ )