Theorems for Calculating Limits

The limit laws reduce complicated limits to simpler ones by breaking them into parts. When these laws cannot be applied directly, because of indeterminate forms, special techniques or theorems such as the Sandwich Theorem are needed.

Law	Formula
Constant Multiple	$lim_{x \to s} [k f (x)] = k lim_{x \to s} f (x)$
Sum	$lim_{x \to s} [f + g] = lim f + lim g$
Product	$lim_{x \to s} [f g] = (lim f) (lim g)$
Quotient	$lim_{x \to s} [f / g] = (lim f) / (lim g)$ , provided $lim g \neq 0$

Throughout this section, $s$ denotes any of $a$ , $a^{+}$ , $a^{-}$ , $- \infty$ , or $+ \infty$ , where $a$ is a real number.

Algebraic Operations on Limits

(Constant Multiple Law) The limit of a constant multiplied by a function is the constant multiplied by the limit of the function: \lim_{x \to s}[kf(x)] = k\lim_{x \to s}f(x). \tag{1}
(Sum Law) The limit of a sum is the sum of the limits (because $f - g = f + (- 1) g$ , the same holds for differences): \lim_{x \to s}[f(x)+g(x)] = \lim_{x \to s}f(x) + \lim_{x \to s}g(x). \tag{2}
If one limit is a number and the other is $+ \infty$ , the result is $+ \infty$ . If one limit is $+ \infty$ and the other is $- \infty$ , the result is an indeterminate form $\infty - \infty$ .
(Product Law) The limit of a product is the product of the limits: \lim_{x \to s}[f(x)g(x)] = \Bigl(\lim_{x \to s}f(x)\Bigr)\Bigl(\lim_{x \to s}g(x)\Bigr). \tag{3}
If one limit is a nonzero number and the other is $\pm \infty$ , the result is $\pm \infty$ (with sign depending on the signs of the factors). If $lim f = 0$ and $lim g = \pm \infty$ , the product $0 \cdot (\pm \infty)$ is an indeterminate form.
(Quotient Law) The limit of a quotient is the quotient of the limits, provided the limit of the denominator is nonzero: \lim_{x \to s}\frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x \to s}f(x)}{\displaystyle\lim_{x \to s}g(x)} \qquad \Bigl(\text{provided } \lim_{x \to s}g(x)\neq 0\Bigr). \tag{4}
If $lim_{x \to s} g (x) = 0$ but $lim_{x \to s} f (x) = L \neq 0$ , then
\lim_{x \to s}\frac{f(x)}{g(x)} = \begin{cases} +\infty & \text{if } L \text{ and } g(x) \text{ have the same signs} \\ -\infty & \text{if } L \text{ and } g(x) \text{ have opposite signs} \\ \text{does not exist} & \text{if the sign of } g(x) \text{ changes} \end{cases}. \tag{5}
If both the numerator and the denominator approach $0$, or both approach $\pm \infty$ , the result is an indeterminate form $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ .

Proof

We prove the Sum Law, Product Law, and Quotient Law for the case

x \to a

(a real number). The Constant Multiple Law is the special case of the Product Law with

g (x) \equiv k

. Before starting, note that

lim_{x \to a} f (x) = L is equivalent to lim_{x \to a} [f (x) - L] = 0.

**Sum Law.** Let

ϵ > 0

be given. We must find

δ > 0

such that

| f (x) + g (x) - L - M | < ϵ

whenever

0 < | x - a | < δ

. By the triangle inequality,

| [f (x) - L] + [g (x) - M] | \leq | f (x) - L | + | g (x) - M |

. Since

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

, there exists

δ_{1} > 0

such that

| f (x) - L | < ϵ / 2

whenever

0 < | x - a | < δ_{1}

. Since

lim_{x \to a} g (x) = M

, there exists

δ_{2} > 0

such that

| g (x) - M | < ϵ / 2

whenever

0 < | x - a | < δ_{2}

. Setting

δ = min {δ_{1}, δ_{2}}

, both conditions hold simultaneously and:

| f (x) + g (x) - L - M | \leq | f (x) - L | + | g (x) - M | < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

whenever

0 < | x - a | < δ

. **Product Law.** We first prove the special case

M = 0

: if

f (x) \to L

and

g (x) \to 0

x \to a

, then

f (x) g (x) \to 0

. Since

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

, there exists

δ_{1} > 0

such that

| f (x) - L | < 1

whenever

0 < | x - a | < δ_{1}

. This means

| f (x) | \leq | f (x) - L | + | L | < 1 + | L |

, so: |f(x)g(x)| = |f(x)||g(x)| < (1+|L|)|g(x)|. \tag{*} Since

lim_{x \to a} g (x) = 0

, there exists

δ_{2} > 0

such that

| g (x) | < \frac{ϵ}{1 + | L |}

whenever

0 < | x - a | < δ_{2}

. \tag{**} Taking

δ = min {δ_{1}, δ_{2}}

, both (*) and (**) hold and

| f (x) g (x) | < ϵ

. For the general case

M \neq 0

, write

f (x) g (x) - L M = [f (x) - L] g (x) + [g (x) - M] L

. Since

f (x) - L \to 0

and

g (x) - M \to 0

, the special case gives

f (x) g (x) - L M \to 0

. **Quotient Law.** It suffices to show

lim_{x \to a} \frac{M}{g (x)} = 1

, since

\frac{f (x)}{g (x)} = \frac{f (x)}{M} \cdot \frac{M}{g (x)}

and the Product Law then gives

\frac{L}{M} \cdot 1 = \frac{L}{M}

. Let

h (x) = g (x) / M

lim_{x \to a} h (x) = 1

. We must show

| \frac{1}{h (x)} - 1 | < ϵ

for

0 < | x - a | < δ

. Since

lim_{x \to a} h (x) = 1

, choose

δ > 0

so that both

| h (x) - 1 | < ϵ / 2

and

| h (x) - 1 | < 1 / 2

when

0 < | x - a | < δ

. The second condition implies

h (x) > 1 / 2

, so

1 / | h (x) | < 2

. Therefore:

| \frac{1}{h (x)} - 1 | = \frac{| h (x) - 1 |}{| h (x) |} < 2 \cdot \frac{ϵ}{2} = ϵ .

Indeterminate Forms

Although the limit laws cover most situations, there are four cases where they give no information:

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

These are called indeterminate forms. Their value cannot be predicted in advance, each may take any value (including $+ \infty$ or $- \infty$ ) or may fail to exist entirely. There are three more indeterminate forms $0^{0}$ , $1^{\pm \infty}$ , and $(\pm \infty)^{0}$ that arise with exponential expressions.

In algebra, $0 / 0$ is simply undefined and $\infty$ is not a number. Here, $0 / 0$ , $0 \cdot (\pm \infty)$ , etc.\ are shorthand for the specific types of indeterminate limits described above.

The Sandwich Theorem

(Sandwich Theorem) Suppose that

g (x) \leq f (x) \leq h (x)

for all $x$ near $a$ (except possibly at $x = a$ itself). If

lim_{x \to a} g (x) = lim_{x \to a} h (x) = L,

then

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

The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.
It also holds for one-sided limits: replace $x \to a$ by $x \to a^{-}$ or $x \to a^{+}$ .

The theorem says: if $f$ is squeezed between $g$ and $h$ near $a$ , and $g$ and $h$ both approach $L$ , then $f$ must also approach $L$ , after all, where else could $f$ go?

Diagram illustrating the Sandwich Theorem: f is squeezed between g and h

Proof

Suppose

ϵ > 0

is given. We need

δ > 0

such that

0 < | x - a | < δ \Rightarrow | f (x) - L | < ϵ

. Because

lim_{x \to a} g (x) = L

, there exists

δ_{1} > 0

such that

0 < | x - a | < δ_{1} \Rightarrow L - ϵ < g (x) < L + ϵ

. Because

lim_{x \to a} h (x) = L

, there exists

δ_{2} > 0

such that

0 < | x - a | < δ_{2} \Rightarrow L - ϵ < h (x) < L + ϵ

. Choose

δ = min {δ_{1}, δ_{2}}

. Then for

0 < | x - a | < δ

L - ϵ < g (x) \leq f (x) \leq h (x) < L + ϵ,

L - ϵ < f (x) < L + ϵ

, i.e.,

| f (x) - L | < ϵ

Show that $lim_{x \to 0} x ⌊ \frac{1}{x} ⌋ = 1$ , where $⌊ \cdot ⌋$ denotes the floor (greatest integer) function.

The floor function satisfies $⌊ x ⌋ =$ the largest integer $\leq x$ , and for every real number $t$ : $t - 1 < ⌊ t ⌋ \leq t$ .

Solution

For every

x \neq 0

\frac{1}{x} - 1 < ⌊ \frac{1}{x} ⌋ \leq \frac{1}{x} .

**If

x > 0

:** multiply by

x

(inequalities preserved):

1 - x < x ⌊ \frac{1}{x} ⌋ \leq 1.

Since

lim_{x \to 0} (1 - x) = 1

, by the Sandwich Theorem:

lim_{x \to 0^{+}} x ⌊ \frac{1}{x} ⌋ = 1.

**If

x < 0

:** multiply by

x

(inequalities reversed):

1 - x > x ⌊ \frac{1}{x} ⌋ \geq 1.

Since

lim_{x \to 0^{-}} (1 - x) = 1

, by the Sandwich Theorem:

lim_{x \to 0^{-}} x ⌊ \frac{1}{x} ⌋ = 1.

Because both one-sided limits equal $1$:

lim_{x \to 0} x ⌊ \frac{1}{x} ⌋ = 1.

Using the inequalities $- | x | \leq \sin x \leq | x |$ and $- | x | \leq 1 - \cos x \leq | x |$ and the Sandwich Theorem, prove:

$lim_{x \to 0} \sin x = 0$
$lim_{x \to 0} \cos x = 1$

Then conclude:

$lim_{x \to a} \sin x = \sin a$
$lim_{x \to a} \cos x = \cos a$

Solution

**(a)** Since

- | x | \leq \sin x \leq | x |

and

lim_{x \to 0} | x | = lim_{x \to 0} (- | x |) = 0

, the Sandwich Theorem gives

lim_{x \to 0} \sin x = 0

. **(b)** Since

- | x | \leq 1 - \cos x \leq | x |

, the Sandwich Theorem gives

lim_{x \to 0} (1 - \cos x) = 0

. Then by the Sum Law,

lim_{x \to 0} (1 - \cos x) = 1 - lim_{x \to 0} \cos x = 0

, so

lim_{x \to 0} \cos x = 1

. **(c)** Let

h = x - a

, so

x \to a

is equivalent to

h \to 0

. Using the Addition Formula for Sine:

lim_{h \to 0} \sin (h + a) = lim_{h \to 0} (\sin h \cos a + \cos h \sin a) = 0 \cdot \cos a + 1 \cdot \sin a = \sin a .

**(d)** Similarly, using the Addition Formula for Cosine:

lim_{h \to 0} \cos (h + a) = lim_{h \to 0} (\cos h \cos a - \sin h \sin a) = 1 \cdot \cos a - 0 \cdot \sin a = \cos a .

Evaluate:

$lim_{x \to 0} x \sin (\frac{1}{x})$
$lim_{x \to 0^{+}} \sqrt{x} \cos (\frac{1}{x^{2}})$
$lim_{x \to 0} \sin x \sin (\frac{1}{x})$

Solution

**(a)** For

x \neq 0

- 1 \leq \sin (1 / x) \leq 1

. For

x > 0

: multiply by

x

to get

- x \leq x \sin (1 / x) \leq x

. Since

lim_{x \to 0^{+}} (\pm x) = 0

, by Sandwich Theorem

lim_{x \to 0^{+}} x \sin (1 / x) = 0

. For

x < 0

: multiply by

x

(inequalities reverse):

- x \geq x \sin (1 / x) \geq x

. Since

lim_{x \to 0^{-}} (\pm x) = 0

, by Sandwich Theorem

lim_{x \to 0^{-}} x \sin (1 / x) = 0

. Both one-sided limits are $0$, so

lim_{x \to 0} x \sin (\frac{1}{x}) = 0

. The graphs of

y = x \sin (1 / x)

y = x

, and

y = - x

are depicted in the following figure.

Graph of y=x sin(1/x) squeezed between y=x and y=-x

**(b)** From

- 1 \leq \cos (1 / x^{2}) \leq 1

, multiply by

\sqrt{x} \geq 0

for

x > 0

- \sqrt{x} \leq \sqrt{x} \cos (\frac{1}{x^{2}}) \leq \sqrt{x} .

Since

lim_{x \to 0^{+}} \sqrt{x} = 0

, by the Sandwich Theorem

lim_{x \to 0^{+}} \sqrt{x} \cos (\frac{1}{x^{2}}) = 0

Graph of y=sqrt(x)cos(1/x^2) squeezed between y=sqrt(x) and y=-sqrt(x)

**(c)** From

- 1 \leq \sin (1 / x) \leq 1

: if

0 < x < π

then

\sin x > 0

, so

- \sin x \leq \sin x \sin (1 / x) \leq \sin x

. If

- π < x < 0

then

\sin x < 0

, so the inequalities reverse. In both cases,

lim_{x \to 0} (\pm \sin x) = 0

, so

lim_{x \to 0} \sin x \sin (\frac{1}{x}) = 0

Graph of y=sin(x)sin(1/x) squeezed between y=sin(x) and y=-sin(x)

The Bounded-Function Theorem

Suppose that $lim_{x \to a} α (x) = 0$ and $f (x)$ is a bounded function near $a$ (but not necessarily at $a$ ): that is, there are numbers $m$ and $M$ such that $m \leq f (x) \leq M$ for all $x$ near $a$ . Then

lim_{x \to a} [α (x) f (x)] = 0.

In words: multiplying a function that approaches zero by a bounded (not blowing-up) function still gives a limit of zero.

For example, since $- 1 \leq \sin (1 / x) \leq 1$ (bounded) and $lim_{x \to 0} x = 0$ :

lim_{x \to 0} x \sin \frac{1}{x} = 0.

Similarly, since $- 1 \leq \sin x \leq 1$ and $lim_{x \to \pm \infty} \frac{1}{x} = 0$ :

lim_{x \to + \infty} \frac{\sin x}{x} = 0, lim_{x \to - \infty} \frac{\sin x}{x} = 0.

Frequently Asked Questions

When can I split a limit into a product or sum?

You can apply the Product Law or Sum Law only when **both** individual limits exist (as finite numbers or

\pm \infty

in compatible combinations). If one limit is infinite and the other is zero, you have an indeterminate form and cannot split directly.

How do I apply the Sandwich Theorem in practice?

Find two simpler functions

g (x)

and

h (x)

that bound

f (x)

from below and above near

a

. Then show both

g

and

h

have the same limit

L

a

. The Sandwich Theorem then guarantees

f (x) \to L

. Bounded oscillating functions like

\sin (1 / x)

are common candidates for sandwiching.

Are there other limit theorems beyond these four laws?

Yes. Additional tools include: (1) the Composite Function Limit Theorem (if

g (x) \to b

and

f

is continuous at

b

, then

lim f (g (x)) = f (b)

), (2) L'Hôpital's Rule for

0 / 0

and

\infty / \infty

forms (covered in the applications of differentiation chapter), and (3) Taylor series methods.

Algebraic Operations on Limits

Indeterminate Forms

The Sandwich Theorem

The Bounded-Function Theorem

Frequently Asked Questions

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