Math Core

Direct substitution handles any limit where plugging in gives a real number. The interesting limits in calculus are exactly the ones where it does not: every derivative you will ever compute starts as a limit of the form $\dfrac{0}{0}$ . This lesson gives you a toolkit of algebraic moves for those limits and a plan for choosing among them.

What substitution tells you

Always start by substituting $x = c$ . The result tells you what to do next.

Result of substituting	What it means	Next step
A real number	The limit is that number (for the functions in this course, at points in their domains).	Done.
$\dfrac{\text{nonzero}}{0}$	The function is unbounded near $c$ ; there is a vertical asymptote.	Check the sign on each side; the limit is $\infty$ , $-\infty$ , or does not exist.
$\dfrac{0}{0}$	Indeterminate form. The limit could be any number, or not exist.	Rewrite the expression algebraically, then substitute again.

Definition

Indeterminate form

A limit is in the indeterminate form $\dfrac{0}{0}$ when the numerator and denominator both approach $0$ . The form itself gives no information about the value of the limit; it only tells you that more work is needed.

The key fact behind every technique below: if two functions agree at every $x$ near $c$ except possibly at $c$ itself, they have the same limit at $c$ . Canceling a factor of $(x - c)$ changes the function only at $x = c$ , which the limit ignores.

Technique 1: Factor and cancel

When a rational function gives $\dfrac{0}{0}$ at $x = c$ , both the numerator and denominator have a factor of $(x - c)$ . Factor, cancel, and substitute.

Worked example: Factoring

Evaluate $\displaystyle \lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 9}$ .

Solution. Substituting gives $\dfrac{9 - 15 + 6}{9 - 9} = \dfrac{0}{0}$ . Factor:

\lim_{x \to 3} \frac{(x - 2)(x - 3)}{(x + 3)(x - 3)} = \lim_{x \to 3} \frac{x - 2}{x + 3} = \frac{1}{6}.

Technique 2: Multiply by the conjugate

When a square root appears in a $\dfrac{0}{0}$ expression, multiply the numerator and denominator by the conjugate. This uses $(a - b)(a + b) = a^2 - b^2$ to clear the root.

Worked example: Rationalizing

Evaluate $\displaystyle \lim_{x \to 4} \frac{\sqrt{2x + 1} - 3}{x - 4}$ .

Solution. Substituting gives $\dfrac{3 - 3}{0} = \dfrac{0}{0}$ . Multiply by $\dfrac{\sqrt{2x + 1} + 3}{\sqrt{2x + 1} + 3}$ :

\begin{aligned} \frac{\sqrt{2x + 1} - 3}{x - 4} \cdot \frac{\sqrt{2x + 1} + 3}{\sqrt{2x + 1} + 3} &= \frac{(2x + 1) - 9}{(x - 4)\left(\sqrt{2x + 1} + 3\right)} \\ &= \frac{2(x - 4)}{(x - 4)\left(\sqrt{2x + 1} + 3\right)} = \frac{2}{\sqrt{2x + 1} + 3}. \end{aligned}

Now substitute: $\dfrac{2}{3 + 3} = \dfrac{1}{3}$ .

Technique 3: Combine fractions

A fraction inside a fraction (a complex fraction) usually simplifies after you combine the small fractions over a common denominator.

Worked example: A complex fraction

Evaluate $\displaystyle \lim_{x \to 0} \frac{\dfrac{1}{x + 4} - \dfrac{1}{4}}{x}$ .

Solution. Substituting gives $\dfrac{0}{0}$ . Combine the fractions in the numerator:

\frac{1}{x + 4} - \frac{1}{4} = \frac{4 - (x + 4)}{4(x + 4)} = \frac{-x}{4(x + 4)}.

Dividing by $x$ cancels the $-x$ (leaving $-1$ ):

\lim_{x \to 0} \frac{-1}{4(x + 4)} = \frac{-1}{16} = -\frac{1}{16}.

Technique 4: Special trigonometric limits

Two limits involving sine and cosine cannot be done by algebra alone. You will prove them in the next lesson using the squeeze theorem; for now, use them as tools. (Angles are in radians.)

Special trigonometric limits

\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \qquad \text{and} \qquad \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0.

More generally, if $u \to 0$ then $\dfrac{\sin u}{u} \to 1$ , whatever the expression $u$ is.

The trick is to make the argument of sine match the denominator. For instance, $\dfrac{\sin 7x}{x} = 7 \cdot \dfrac{\sin 7x}{7x}$ , and as $x \to 0$ we have $7x \to 0$ , so the limit is $7 \cdot 1 = 7$ .

Worked example: A trigonometric limit

Evaluate $\displaystyle \lim_{x \to 0} \frac{\tan 3x}{\sin 2x}$ .

Solution. Substituting gives $\dfrac{0}{0}$ . Write $\tan 3x = \dfrac{\sin 3x}{\cos 3x}$ and insert matching factors:

\frac{\tan 3x}{\sin 2x} = \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{1}{\cos 3x} \cdot \frac{3x}{2x}.

As $x \to 0$ , the first two factors approach $1$ , the third approaches $\dfrac{1}{\cos 0} = 1$ , and the last one equals $\dfrac{3}{2}$ for every $x \ne 0$ . The limit is $\dfrac{3}{2}$ .

Absolute values and piecewise functions

When the expression contains $\lvert x - c \rvert$ , the formula changes at $c$ , so find the one-sided limits separately. For example, $\lvert x - 4 \rvert = x - 4$ when $x > 4$ and $\lvert x - 4 \rvert = -(x - 4)$ when $x < 4$ . So

\lim_{x \to 4^+} \frac{\lvert x - 4 \rvert}{x - 4} = 1, \qquad \lim_{x \to 4^-} \frac{\lvert x - 4 \rvert}{x - 4} = -1,

and the two-sided limit does not exist.

Common mistake

Writing $\displaystyle \lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 9} = \frac{0}{0}$ and stopping, or saying "the limit is undefined," is wrong. The form $\dfrac{0}{0}$ is a starting point, not an answer. Also, keep writing " $\displaystyle \lim_{x \to c}$ " on every line until you actually substitute: the expression before substitution is not equal to its limit.

Tip

Choosing a technique: a polynomial ratio means factor; a square root means conjugate; fractions inside fractions mean combine; sine, cosine or tangent means special trig limits; an absolute value means check both sides. On the AP exam you can check your algebraic answer against a quick table of values.