Determining Limits Using Algebraic Manipulation (AP Calculus AB 1.6)

Substituting $x = 5$ into $\dfrac{x^2 - 25}{x - 5}$ gives $\frac{0}{0}$ , which is not a number. Yet for every $x \neq 5$ the fraction equals $x + 5$ , so its graph is the line $y = x + 5$ with a single point missing at $(5, 10)$ , and the outputs plainly approach 10. Every method in this topic does what that one line of factoring did: rewrite the expression into a form that agrees with it everywhere except at the troublesome point, then substitute into the new form.

What $\frac{0}{0}$ does and doesn't tell you

When both the numerator and the denominator approach 0, the quotient law from topic 1.5 can't be used, and the limit could turn out to be anything. Near $x = 0$ :

Expression	Simplifies to (for $x \neq 0$ )	Limit as $x \to 0$
$\dfrac{x}{x}$	$1$	$1$
$\dfrac{3x}{x}$	$3$	$3$
$\dfrac{x^2}{x}$	$x$	$0$
$\dfrac{x}{x^2}$	$\dfrac{1}{x}$	does not exist

All four start as $\frac{0}{0}$ and end in four different places, because what matters is how fast the top and bottom shrink compared with each other. That is why $\frac{0}{0}$ is called an indeterminate form: it means "rewrite and look again," never "the answer is 0," "the answer is 1," or "the limit doesn't exist."

A nonzero number over something approaching 0 is a different situation. The quotient's size grows without bound, so there is no finite limit, and you decide between $\infty$ , $-\infty$ and neither by checking signs on each side (topic 1.14). Algebra won't change that outcome.

The reason rewriting is allowed is that a limit at $a$ never looks at $x = a$ itself, only at $x$ near $a$ .

Canceling $(x - 5)$ from $\frac{x^2 - 25}{x - 5}$ is legal only for $x \neq 5$ , and that is exactly the condition the box asks for.

Factoring and canceling

If a polynomial $p$ has $p(a) = 0$ , then $x - a$ is a factor of $p$ (the factor theorem). So when a quotient of polynomials gives $\frac{0}{0}$ at $x = a$ , both the top and the bottom contain the factor $x - a$ , and canceling it is the whole job. The factorings you need most:

trinomials: $x^2 + x - 6 = (x + 3)(x - 2)$ ;
difference of squares: $a^2 - b^2 = (a - b)(a + b)$ ;
difference and sum of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ .

After one cancellation, substitute again. Sometimes the result is still $\frac{0}{0}$ , and you factor again. Sometimes only the numerator is still 0, and then the limit is 0: $\dfrac{x^2 - 2x + 1}{x^2 - 1} = \dfrac{(x - 1)^2}{(x - 1)(x + 1)} = \dfrac{x - 1}{x + 1}$ for $x \neq 1$ , which approaches $\frac{0}{2} = 0$ as $x \to 1$ .

When the expression isn't factored yet, expand it first. With $h$ as the variable,

\lim_{h \to 0}\frac{(3 + h)^2 - 9}{h} = \lim_{h \to 0}\frac{6h + h^2}{h} = \lim_{h \to 0}(6 + h) = 6.

Limits shaped like this one become the definition of the derivative in topic 2.2, so expanding and canceling an $h$ is worth making automatic.

Multiplying by a conjugate

A square root in a difference, as in $\sqrt{9 + x} - 3$ , can't be factored with integer patterns. The fix is the difference of squares in reverse: $(A - B)(A + B) = A^2 - B^2$ squares the root away. The expression $A + B$ is called the conjugate of $A - B$ . Multiply the top and bottom by it:

\lim_{x \to 0}\frac{\sqrt{9 + x} - 3}{x}\cdot\frac{\sqrt{9 + x} + 3}{\sqrt{9 + x} + 3} = \lim_{x \to 0}\frac{(9 + x) - 9}{x\left(\sqrt{9 + x} + 3\right)} = \lim_{x \to 0}\frac{1}{\sqrt{9 + x} + 3} = \frac{1}{6}.

Leave the conjugate in the denominator unmultiplied. You want the factor $x$ to appear on top so it can cancel, and expanding the bottom would only hide it. The same move works when the root is in the denominator; multiply by the denominator's conjugate instead.

Combining fractions

A fraction inside a fraction usually hides the cancellation. Combine the small fractions over a common denominator first:

\lim_{x \to 3}\frac{\frac{3}{x} - 1}{x - 3} = \lim_{x \to 3}\frac{\frac{3 - x}{x}}{x - 3} = \lim_{x \to 3}\frac{-(x - 3)}{x(x - 3)} = \lim_{x \to 3}\left(-\frac{1}{x}\right) = -\frac{1}{3}.

Watch for $3 - x = -(x - 3)$ : it is the same factor with the sign flipped, and missing it flips the sign of the answer.

Trig identities

For trig expressions that give $\frac{0}{0}$ , an identity can produce the common factor. The ones that do the most work:

$\sin^2 x + \cos^2 x = 1$ , so $1 - \cos^2 x = \sin^2 x$ and $1 - \sin^2 x = \cos^2 x$ ;
$\sin 2x = 2\sin x\cos x$ ;
$\cos 2x = \cos^2 x - \sin^2 x$ , which factors as $(\cos x - \sin x)(\cos x + \sin x)$ ;
$\tan x = \dfrac{\sin x}{\cos x}$ .

For example, $\dfrac{\sin 2x}{\sin x} = \dfrac{2\sin x\cos x}{\sin x} = 2\cos x$ whenever $\sin x \neq 0$ , so $\lim\limits_{x \to 0}\dfrac{\sin 2x}{\sin x} = 2\cos 0 = 2$ .

Some trig limits have no factor to cancel. $\lim\limits_{x \to 0}\frac{\sin x}{x}$ is $\frac{0}{0}$ , but no identity turns it into something you can substitute into; it takes the squeeze theorem of topic 1.8, and topic 1.7 is about recognizing which method a limit needs.

Worked examples

Worked exampleExample 1: factoring a sum of cubes

Find $\lim\limits_{x \to -2}\dfrac{x^3 + 8}{x^2 + x - 2}$ .

Show solution

Substitution gives $\frac{-8 + 8}{4 - 2 - 2} = \frac{0}{0}$ , so both polynomials have the factor $x + 2$ . Factor the top as a sum of cubes with $a = x$ , $b = 2$ , and the bottom as a trinomial:

\frac{x^3 + 8}{x^2 + x - 2} = \frac{(x + 2)(x^2 - 2x + 4)}{(x + 2)(x - 1)} = \frac{x^2 - 2x + 4}{x - 1} \quad\text{for } x \neq -2.

The new denominator is $-3$ at $x = -2$ , so substitution works:

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

Worked exampleExample 2: a conjugate, then a factor

Find $\lim\limits_{x \to 4}\dfrac{\sqrt{2x + 1} - 3}{x^2 - 16}$ .

Show solution

Substitution gives $\frac{3 - 3}{0} = \frac{0}{0}$ . The root sits in a difference on top, so multiply by its conjugate:

\frac{\sqrt{2x + 1} - 3}{x^2 - 16}\cdot\frac{\sqrt{2x + 1} + 3}{\sqrt{2x + 1} + 3} = \frac{(2x + 1) - 9}{(x^2 - 16)\left(\sqrt{2x + 1} + 3\right)} = \frac{2(x - 4)}{(x - 4)(x + 4)\left(\sqrt{2x + 1} + 3\right)}.

The numerator became $2x - 8 = 2(x - 4)$ , and factoring $x^2 - 16$ shows the matching factor. Cancel it (allowed because $x \neq 4$ ) and substitute:

\lim_{x \to 4}\frac{2}{(x + 4)\left(\sqrt{2x + 1} + 3\right)} = \frac{2}{8 \cdot 6} = \frac{1}{24}.

Worked exampleExample 3: two trig limits

Find (a) $\lim\limits_{x \to \pi/4}\dfrac{\cos 2x}{\cos x - \sin x}$ and (b) $\lim\limits_{x \to 0}\dfrac{1 - \cos x}{\sin^2 x}$ .

Show solution

(a) At $x = \frac{\pi}{4}$ , $\cos 2x = \cos\frac{\pi}{2} = 0$ and $\cos x = \sin x = \frac{\sqrt{2}}{2}$ , so this is $\frac{0}{0}$ . The double-angle identity factors the top:

\frac{\cos 2x}{\cos x - \sin x} = \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x} = \cos x + \sin x,

so the limit is $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$ .

(b) At $x = 0$ this is $\frac{1 - 1}{0} = \frac{0}{0}$ . Write $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$ , which shows the common factor:

\frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} = \frac{1}{1 + \cos x}, \qquad \lim_{x \to 0}\frac{1}{1 + \cos x} = \frac{1}{2}.

Worked exampleExample 4: a constant that makes the limit exist (exam style, no calculator)

Let $f(x) = \dfrac{x^2 + kx - 12}{x - 2}$ for $x \neq 2$ , where $k$ is a constant.

(a) For what value of $k$ does $\lim\limits_{x \to 2} f(x)$ exist? Justify your answer.

(b) For the value of $k$ found in part (a), find $\lim\limits_{x \to 2} f(x)$ .

Show solution

(a) The denominator approaches 0 as $x \to 2$ . If the numerator approached a nonzero number, the quotient's size would grow without bound and the limit could not exist. So the numerator must approach 0:

2^2 + 2k - 12 = 0 \quad\Longrightarrow\quad 2k = 8 \quad\Longrightarrow\quad k = 4.

That condition is necessary but not yet a proof the limit exists; part (b) finishes the job by finding it.

(b) With $k = 4$ , the numerator has $x - 2$ as a factor: $x^2 + 4x - 12 = (x - 2)(x + 6)$ . For $x \neq 2$ , $f(x) = x + 6$ , so

\lim_{x \to 2} f(x) = \lim_{x \to 2}(x + 6) = 8.

(c) With $k = 1$ , the numerator approaches $4 + 2 - 12 = -6$ while the denominator approaches 0. For $x$ slightly less than 2, $x - 2$ is a small negative number, and $\frac{\text{negative}}{\text{small negative}}$ is a large positive number, so $\lim\limits_{x \to 2^-} f(x) = \infty$ . For $x$ slightly more than 2, the denominator is a small positive number, so $\lim\limits_{x \to 2^+} f(x) = -\infty$ .

Where students lose points

On the exam

In multiple choice without a calculator, these rewrites are what you use when substitution gives $\frac{0}{0}$ and you are asked for the limit's value. In free response, they can come up inside other questions: finding a constant that makes a function continuous (topic 1.13), or evaluating a limit in a definition of the derivative.

Check yourself

Question 1
The function $f$ satisfies $f(x) = x^2 + 1$ for every $x \neq 3$ , and $f(3) = 0$ . What is $\lim\limits_{x \to 3} f(x)$ ?
1. A
  $10$
2. B
  $0$
3. C
  The limit does not exist.
4. D
  It cannot be determined from the information given.
Show answer
A
The limit at 3 only uses values of $f$ near 3, where $f(x) = x^2 + 1$ , so it equals $\lim\limits_{x \to 3}(x^2 + 1) = 10$ . Choice B uses the value $f(3)$ , which a limit never looks at.
Question 2
Find $\lim\limits_{x \to 0}\dfrac{x}{\sqrt{x + 4} - 2}$ .
Show answer
$4$
Multiply top and bottom by the denominator's conjugate, $\sqrt{x + 4} + 2$ . The bottom becomes $(x + 4) - 4 = x$ , which cancels the $x$ on top, leaving $\sqrt{x + 4} + 2 \to 2 + 2 = 4$ .
Question 3
What is $\lim\limits_{x \to \pi/2}\dfrac{\cos x}{1 - \sin^2 x}$ ?
1. A
  $0$
2. B
  $1$
3. C
  $\dfrac{1}{2}$
4. D
  The limit does not exist.
Show answer
D
It starts as $\frac{0}{0}$ . Since $1 - \sin^2 x = \cos^2 x$ , the expression equals $\frac{1}{\cos x}$ , which is $\frac{1}{0}$ at $\frac{\pi}{2}$ : it goes to $\infty$ from the left and $-\infty$ from the right. Rewriting doesn't always produce a number; after a cancellation, substitute again and read what you get.
Question 4
The constants $a$ and $b$ satisfy $\lim\limits_{x \to 1}\dfrac{x^2 + ax + b}{x - 1} = 5$ . Find $a$ and $b$ .
Show answer
$a = 3$ and $b = -4$
The denominator approaches 0 and the limit is finite, so the numerator must approach 0: $1 + a + b = 0$ . Then $x - 1$ is a factor, $x^2 + ax + b = (x - 1)(x + c)$ , and the limit is $1 + c = 5$ , so $c = 4$ . Expanding $(x - 1)(x + 4) = x^2 + 3x - 4$ gives $a = 3$ , $b = -4$ .

Go deeper

OpenStax, Calculus Volume 1, 2.3 The Limit Laws

Determining Limits Using Algebraic Manipulation

What 00\frac{0}{0}00​ does and doesn't tell you

Factoring and canceling

Multiplying by a conjugate

Combining fractions

Trig identities

Worked examples

Where students lose points

On the exam

Check yourself

Go deeper

What $\frac{0}{0}$ does and doesn't tell you