Limits by Algebraic Manipulation

The theorem holds because a limit never looks at

x = a

itself. When substituting

a

gives

\frac{0}{0}

, called an indeterminate form, the substitution names no value, and the limit could be any number or fail to exist. Each method on this page rewrites the expression, for

x

near

a

with

x \ne a

, as one that can be evaluated by substitution. With an absolute value the rewrite is different on each side of

a

. The figure shows the simplest case.

y = (x² − 2x − 8)/(x + 2) is the line y = x − 4 without the point (−2, −6). Its limit at −2 is −6.

Factoring

Example 1

Find each limit.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 2} \frac{x^3 + 2x^2 - 4x - 8}{x^2 - 4} \\[10pt] &\text{(b)}\;\; \lim_{x \to -1} \frac{x^3 + 1}{x^2 - 1} \\[10pt] &\text{(c)}\;\; \lim_{h \to 0} \frac{(h - 2)^2 - 4}{h} \end{aligned}

(a) Substitution gives

\frac{0}{0}

, so

x - 2

is a factor of both the numerator and the denominator. Factor the numerator by grouping:

\begin{aligned} &x^3 + 2x^2 - 4x - 8 \\[4pt] &\qquad = x^2(x + 2) - 4(x + 2) \\[4pt] &\qquad = (x + 2)^2(x - 2). \end{aligned}

For

x \ne 2

the factor

x - 2

cancels, so

\begin{aligned} &\lim_{x \to 2} \frac{x^3 + 2x^2 - 4x - 8}{x^2 - 4} \\[4pt] &\qquad = \lim_{x \to 2} \frac{(x + 2)^2(x - 2)}{(x + 2)(x - 2)} \\[4pt] &\qquad = \lim_{x \to 2} (x + 2) = 2 + 2 = 4. \end{aligned}

(b) The numerator is a sum of cubes,

x^3 + 1 = (x + 1)(x^2 - x + 1)

, and the denominator is

(x + 1)(x - 1)

. For

x \ne -1

the quotient is

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

, so

\begin{aligned} &\lim_{x \to -1} \frac{x^2 - x + 1}{x - 1} \\[4pt] &\qquad = \frac{(-1)^2 - (-1) + 1}{-1 - 1} = -\frac{3}{2}. \end{aligned}

(c) Expand first. The numerator is

h^2 - 4h + 4 - 4 = h(h - 4)

, so for

h \ne 0

the quotient is

h - 4

, and the limit is

-4

Rationalizing

When a square root causes the

\frac{0}{0}

, multiply the numerator and the denominator by the conjugate:

\sqrt{A} + B

for

\sqrt{A} - B

. The product

(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2

has no square root left.

Example 2

Find each limit.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2} \\[10pt] &\text{(b)}\;\; \lim_{x \to 0} \frac{x^2 - 3x}{3 - \sqrt{9 + x}} \end{aligned}

(a) Multiply by

\sqrt{x + 7} + 3

above and below. The numerator becomes

(x + 7) - 9 = x - 2

, which cancels:

\begin{aligned} &\frac{\sqrt{x + 7} - 3}{x - 2} \\[4pt] &\qquad = \frac{x - 2}{(x - 2)\left(\sqrt{x + 7} + 3\right)} \\[4pt] &\qquad = \frac{1}{\sqrt{x + 7} + 3} \end{aligned}

for

x \ne 2

. The limit is

\frac{1}{\sqrt{9} + 3} = \frac{1}{6}

(b) The root is in the denominator, so multiply by its conjugate,

3 + \sqrt{9 + x}

. The denominator becomes

9 - (9 + x) = -x

, and for

x \ne 0

\begin{aligned} &\frac{x^2 - 3x}{3 - \sqrt{9 + x}} \\[4pt] &\qquad = \frac{x(x - 3)\left(3 + \sqrt{9 + x}\right)}{-x} \\[4pt] &\qquad = -(x - 3)\left(3 + \sqrt{9 + x}\right). \end{aligned}

The limit is

-(0 - 3)\left(3 + \sqrt{9}\right) = 18

Combining fractions

Absolute values

When the inside of an absolute value changes sign at

a

, the absolute value is one expression on the left of

a

and another on the right. Find the two one-sided limits separately.

Example 4

Let

f(x) = \dfrac{|x - 3|}{x^2 - 3x}

. Find each limit, or explain why it doesn't exist.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 3^-} f(x) \\[8pt] &\text{(b)}\;\; \lim_{x \to 3^+} f(x) \\[8pt] &\text{(c)}\;\; \lim_{x \to 3} f(x) \end{aligned}

The denominator factors as

x(x - 3)

(a) For

0 < x < 3

x - 3

is negative, so

|x - 3| = -(x - 3)

and

f(x) = \frac{-(x - 3)}{x(x - 3)} = -\frac{1}{x}.

The left-hand limit is

-\frac{1}{3}

(b) For

x > 3

|x - 3| = x - 3

, so

f(x) = \frac{1}{x}

, and the right-hand limit is

\frac{1}{3}

(c) The one-sided limits differ, so

\lim_{x \to 3} f(x)

doesn't exist.

Near 3, the graph of f is y = −1/x on the left and y = 1/x on the right.

Trig identities

Example 5

Find each limit.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to \pi} \frac{\tan^2 x}{1 + \sec x} \\[10pt] &\text{(b)}\;\; \lim_{x \to \pi} \frac{\sin 2x}{\sin x} \end{aligned}

(a) Since

\tan \pi = 0

and

\sec \pi = -1

, substitution gives

\frac{0}{0}

. Factor the numerator with the identity

\tan^2 x = \sec^2 x - 1

\begin{aligned} \tan^2 x &= \sec^2 x - 1 \\[4pt] &= (\sec x - 1)(\sec x + 1). \end{aligned}

For

x

near

\pi

with

x \ne \pi

\sec x \ne -1

, so the factor

\sec x + 1

cancels and the quotient is

\sec x - 1

. The limit is

\sec \pi - 1 = -2

Between π/2 and 3π/2, y = tan²x/(1 + sec x) is y = sec x − 1 without the point (π, −2).

(b) By the double-angle identity,

\sin 2x = 2\sin x\cos x

. For

x

near

\pi

with

x \ne \pi

\sin x \ne 0

, and the quotient is

2\cos x

. The limit is

2\cos\pi = -2

Choosing a method

First check whether the function is piecewise with a rule that changes at

a

. If it isn't, substitute

a

. What comes out tells you the next step.

What you see	Next step
A piecewise rule that changes at $a$	Find the two one-sided limits, each from its own piece.
One formula, and substituting $a$ gives a number	That's the limit, when the formula is built from polynomials, roots, trig, exponential, and log functions and is defined on an open interval around $a$ .
Substituting $a$ gives $\dfrac{b}{0}$ with $b \ne 0$	The limit doesn't exist. Check the sign on each side: when both sides agree, write $\infty$ or $-\infty$ .
Substituting $a$ gives $\dfrac{0}{0}$	Factor, rationalize, combine fractions, or use an identity. Split an absolute value by sides.

Example 6

Find each limit, or explain why it doesn't exist. In (d),

f

is the function below the list.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 1} \frac{x^2 + 3}{x + 1} \\[10pt] &\text{(b)}\;\; \lim_{x \to 1} \frac{x + 3}{(x - 1)^2} \\[10pt] &\text{(c)}\;\; \lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9} \\[10pt] &\text{(d)}\;\; \lim_{x \to 2} f(x) \end{aligned}

f(x) = \begin{cases} x^2 + 1, & x < 2 \\ 3x - 1, & x \ge 2 \end{cases}

(a) Substitution gives

\frac{4}{2} = 2

, and the denominator isn't

0

, so the limit is

2

(b) Substitution gives

\frac{4}{0}

. The denominator

(x - 1)^2

is positive on both sides of

1

and approaches

0

, while the numerator approaches

4

. So the limit doesn't exist: the quotient grows without bound on both sides, written

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

(c) Substitution gives

\frac{0}{0}

. Factor the difference of cubes and the difference of squares:

\frac{x^3 - 27}{x^2 - 9} = \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)}.

For

x \ne 3

the quotient is

\frac{x^2 + 3x + 9}{x + 3}

, so

\lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9} = \frac{27}{6} = \frac{9}{2}

(d) The rule changes at

2

. From the left,

x^2 + 1

approaches

5

. From the right,

3x - 1

approaches

5

. The one-sided limits agree, so

\lim_{x \to 2} f(x) = 5