4.4 Theorems for Calculating Limits

Estimating the limit of a function using the graphical approach may not be very accurate, and as we saw in Example 4 of Section 4.1, the numerical approach may lead to incorrect results. In this section, we discuss how we can evaluate limits analytically.

In this section, we learn:

Table of Contents

Algebraic Operations on Limits

We first begin with some useful limits:

Theorem 1: 1. (Constant function rule): If $c$ is a constant, then for any number $a$

$lim_{x \to a} c = c$

2. (Identity function rule):

$lim_{x \to a} x = a .$

The Constant Function Rule says that if $f (x)$ is identically equal to $c$ , then $f (x)$ is close to $c$ for all $x$ close to $a$ . This is true because $f (x) = c$ is always close to $c$ .

The Identity Function Rule merely says that $f (x) = x$ is close to $a$ whenever $x$ is close to $a$ .

The illustrations of these two rules are shown in Figure 1.


(a) Graph of a constant function. It is clear that $lim_{x \to a} c = c$	(b) Graph of the identity function $y = x$ . It is clear that $lim_{x \to a} x = a$ .

Figure 1

Although the above limits are obvious from the intuitive viewpoint, we can prove them rigorously using the $ϵ$ – $δ$ definition.

Show the Proof of Theorem 1

(1) Constant function rule: Here $f (x) = c$ for every $x$ and $L = c$ . We must prove that for every $ϵ > 0$ , there exists a $δ > 0$ such that

$0 < | x - a | < δ ⟹ | c - c | < ϵ$

Because $| c - c | = 0$ , we always have

$| c - c | < ϵ .$

So we can choose any positive number for $δ$ , and get

$| c - c | < ϵ whenever 0 < | x - a | < δ$

(2) Identity function rule: Here $f (x) = x$ and $L = a$ . We must prove that for every $ϵ > 0$ , there exists a $δ > 0$ such that

$0 < | x - a | < δ ⟹ | x - a | < ϵ$

Obviously if we choose $δ \leq ϵ$ , then

$0 < | x - a | < δ ⟹ | x - a | < δ \leq ϵ .$

Basic Algebraic Operations

We can evaluate many limits by applying the following limit laws.

Theorem 2. (Algebraic Operations on Limits): Let

a

k

L

, and

M

be real numbers. If

f

and

g

are functions such that

$lim_{x \to a} f (x) = L, and lim_{x \to a} g (x) = M$

then

1. Constant multiple rule:

$lim_{x \to a} [k f (x)] = k L .$

2. Sum rule:

$lim_{x \to a} [f (x) + g (x)] = L + M$

3. Difference rule:

$lim_{x \to a} [f (x) - g (x)] = L - M$

4. Product rule:

$lim_{x \to a} [f (x) g (x)] = L M$

5. Quotient rule:

$lim_{x \to a} \frac{f (x)}{g (x)} = \frac{L}{M} (provided M \neq 0)$

Similar results hold for left and right hand limits; that is, we can replace $x \to a$ by $x \to a^{-}$ or $x \to a^{+}$ in the above equations.

This theorem merely says:

The limit of a constant times a function is the constant times the limit of the function.
The limit of a sum is the sum of the limits.
The limit of a difference is the difference of the limits.
The limit of a product is the product of the limits.
The limit of a quotient is the quotient of the limits provided that the limit of the denominator is not zero.

Note that the Constant Multiple Rule (1) is a special case of the Product Rule (4) when $g$ is a constant function, namely $g (x) = k$ .

This theorem comes from the common sense that if the number $u_{1}$ is close to the number $v_{1}$ and the number $u_{2}$ is close to the number $v_{2}$ then $u_{1} \pm u_{2}$ will be close to $v_{1} \pm v_{2}$ , $u_{1} u_{2}$ will be close to $v_{1} v_{2}$ , and $u_{1} / u_{2}$ will be close to $v_{1} / v_{2}$ . For example,

$\begin{aligned} 7.01 + 11.001 & \approx 7 + 11 \\ 7.01 - 11.001 & \approx 7 - 11 \\ 7.01 \times 11.001 & \approx 7 \times 11 \\ 7.01 / 11.001 & \approx 7 / 11 \end{aligned}$

So it is easy to believe if $f (x)$ is close to $L$ and $g (x)$ is close to $M$ when $x$ is close to $a$ , then $f (x) + g (x)$ is close to $L + M$ , and $f (x) g (x)$ is close to $L M$ when $x$ is close to $a$ . Similarly $f (x) / g (x)$ will be close to $L / M$ if $M \neq 0$ . In the Quotient Rule, we have to exclude the case of $M = 0$ because division by zero is not defined.

Using the $ϵ$ – $δ$ definition, we can prove this theorem with mathematical rigor.
The above theorem can be extended to the sum, difference, or product of any finite number of functions. Namely if
$lim_{x \to a} f_{1} (x) = L_{1}, lim_{x \to a} f_{2} (x) = L_{2}, \dots, lim_{x \to f_{n} (x)} = L_{n}$ then
$lim_{x \to a} [f_{1} (x) \pm f_{2} (x) \pm \dots \pm f_{n} (x)] = L_{1} \pm L_{2} \pm \dots \pm L_{n}$ and
$lim_{x \to a} [f_{1} (x) f_{2} (x) \dots f_{n} (x)] = L_{1} L_{2} \dots L_{n}$ Specifically, if $n$ is a positive integer then
$lim_{x \to a} {[f (x)]}^{n} = L^{n} .$

Example 1

The graphs of $f$ and $g$ are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist.

(a) $lim_{x \to - 3} [3 f (x) - 2 g (x)]$

(b) $lim_{x \to - 3} \frac{f (x)}{4 g (x)}$

Solution

(a) By the Sum Rule we have

$lim_{x \to - 3} [3 f (x) - 2 g (x)] = lim_{x \to - 3} [3 f (x)] + lim_{x \to - 3} [- 2 g (x)]$

By the Constant Multiple Rule we have

$lim_{x \to - 3} [3 f (x)] + lim_{x \to - 3} [- 2 g (x)] = 3 lim_{x \to - 3} f (x) - 2 lim_{x \to - 3} g (x)$

From the graphs of $f$ and $g$ we find that

$lim_{x \to - 3} f (x) = 3, and lim_{x \to - 3} g (x) = - 2$

Therefore:

$\begin{aligned} lim_{x \to - 3} [3 f (x) - 2 g (x)] = & 3 lim_{x \to - 3} f (x) - 2 lim_{x \to - 3} g (x) \\ = & 3 \times 3 - 2 \times (- 2) = 13. \end{aligned}$

(b) By the Constant Multiple Rule:

$lim_{x \to - 3} \frac{f (x)}{4 g (x)} = \frac{1}{4} lim_{x \to - 3} \frac{f (x)}{g (x)}$

and by the Quotient Rule:

$\frac{1}{4} lim_{x \to - 3} \frac{f (x)}{g (x)} = \frac{1}{4} \frac{lim_{x \to - 3} f (x)}{lim_{x \to - 3} g (x)}$

Because

$lim_{x \to - 3} f (x) = 3 and lim_{x \to - 3} g (x) = - 2,$

we get

$lim_{x \to - 3} \frac{f (x)}{4 g (x)} = \frac{1}{4} \frac{lim_{x \to - 3} f (x)}{lim_{x \to - 3} g (x)} = \frac{1}{4} \frac{3}{- 2} = - \frac{3}{8} .$

(c) From the graphs of $f$ and $g$ , we see $lim_{x \to 0} f (x) = - 2$ , but $lim_{x \to 0} g (x)$ does not exist, because the left and right limits are not the same

$lim_{x \to 0^{-}} g (x) = 2, lim_{x \to 0^{+}} g (x) = 3.$

By the Product Rule the left-hand limit is

$lim_{x \to 0^{-}} [f (x) g (x)] = (lim_{x \to 0^{-}} f (x)) (lim_{x \to 0^{-}} g (x)) = - 2 \times 2 = - 4$

and the right-hand limit is:

$lim_{x \to 0^{+}} [f (x) g (x)] = (lim_{x \to 0^{+}} f (x)) (lim_{x \to 0^{+}} g (x)) = - 2 \times 3 = - 6.$

Because the left and right limits are not equal, $lim_{x \to 0} [f (x) g (x)]$ does not exist.

Example 2

The graphs of $f$ and $g$ are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist.

(a) $lim_{x \to 1} [f (x) - g (x)]$

(b) $lim_{x \to 1} [f (x) g (x)]$

Solution

As $lim_{x \to 1} g (x)$ does not exist, following the previous example, we may be tempted to say that none of the above limits exist. However, we will see that only the first one does not exist.

(a)

$\begin{aligned} lim_{x \to 1^{-}} [f (x) - g (x)] & = lim_{x \to 1^{-}} f (x) - lim_{x \to 1^{-}} g (x) \\ = 2 - 0 = 0. \end{aligned}$

$\begin{aligned} lim_{x \to 1^{+}} [f (x) - g (x)] & = lim_{x \to 1^{+}} f (x) - lim_{x \to 1^{+}} g (x) \\ = - 3 - 0 = - 3 \end{aligned}$

By Theorem 1 in Section 4.3, we know that the two-sided limit exists if and only if the one-sided limits exist and are equal. So because

$lim_{x \to 1^{-}} [f (x) - g (x)] \neq lim_{x \to 1^{+}} [f (x) - g (x)],$

we conclude that $lim_{x \to 1} [f (x) - g (x)]$ does not exist.

(b)

$\begin{aligned} lim_{x \to 1^{-}} [f (x) g (x)] & = (lim_{x \to 1^{-}} f (x)) (lim_{x \to 1^{-}} g (x)) \\ = 2 \times 0 \\ = 0. \end{aligned}$

$\begin{aligned} lim_{x \to 1^{+}} [f (x) g (x)] & = (lim_{x \to 1^{+}} f (x)) (lim_{x \to 1^{+}} g (x)) \\ = - 3 \times 0 \\ = 0. \end{aligned}$

Because the left and right hand limits are both equal to 0, we conclude that

$lim_{x \to 1} [f (x) g (x)] = 0$

(c)

$\begin{aligned} lim_{x \to 1^{-}} \frac{g (x)}{f (x)} & = \frac{lim_{x \to 1^{-}} g (x)}{lim_{x \to 1^{-}} f (x)} \\ = \frac{0}{2} = 0, \end{aligned}$

and

$\begin{aligned} lim_{x \to 1^{+}} \frac{g (x)}{f (x)} & = \frac{lim_{x \to 1^{+}} g (x)}{lim_{x \to 1^{+}} f (x)} \\ = \frac{0}{- 3} = 0. \end{aligned}$

Here because the left and right hand limits exist and equal, we conclude that the two sided limit exists too and its value is 0:

$lim_{x \to 1} \frac{g (x)}{f (x)} = 0.$

Example 3

Show that $lim_{x \to 0} | x | = 0$ .

Solution

We note that

$| x | = {\begin{cases} x & if x \geq 0 \\ - x & if x < 0 \end{cases}$

The right-hand limit is

$lim_{x \to 0^{+}} | x | = lim_{x \to 0^{+}} x = 0,$

and the left-hand limit is

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

Because

$lim_{x \to 0^{+}} | x | = lim_{x \to 0^{-}} | x | = 0$

It follows from Theorem 1 in Section 4.3 that

$lim_{x \to 0} | x | = 0.$

Limits of Polynomial and Rational Functions

Example 4

Evaluate $lim_{x \to 2} (4 x^{2} - 3 x + 1)$ .

Solution

$\begin{aligned} lim_{x \to 2} (4 x^{2} - 3 x + 1) = & (lim_{x \to 2} (4 x^{2})) - (lim_{x \to 2} (3 x)) + (lim_{x \to 2} 1) \\ = & 4 (lim_{x \to 2} x^{2}) - 3 (lim_{x \to 2} x) + 1 \\ = & 4 (lim_{x \to 2} x) (lim_{x \to 2} x) - 3 \times 2 + 1 \\ = & 4 \times 2 \times 2 - 3 \times 2 + 1 \\ = & 16 - 6 + 1 = 11. \end{aligned}$

In this example, it is as if we just substituted 2 for $x$ in the polynomial $4 x^{2} - 3 x + 1$ .

Example

Evaluate

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

Solution

$\begin{aligned} lim_{x \to - 1} \frac{4 x^{2} - 3 x}{5 x^{3} + 1} & = \frac{lim_{x \to - 1} (4 x^{2} - 3 x)}{lim_{x \to - 1} (5 x^{3} + 1)} \\ = \frac{[lim_{x \to - 1} (4 x^{2})] - [lim_{x \to - 1} (3 x)]}{[lim_{x \to - 1} (5 x^{3})] + lim_{x \to - 1} 1} \\ = \frac{4 [lim_{x \to - 1} (x^{2})] - 3 lim_{x \to - 1} x}{5 [lim_{x \to - 1} (x^{3})] + 1} \\ = \frac{4 {(lim_{x \to - 1} x)}^{2} - 3 lim_{x \to - 1} x}{5 {(lim_{x \to - 1} x)}^{3} + 1} \\ = \frac{4 (- 1)^{2} - 3 (- 1)}{5 (- 1)^{3} + 1} \\ = - \frac{7}{4} \end{aligned}$

In general:

Theorem 3. (Limits of Polynomials and Rational Functions): Let $P (x)$ and $Q (x)$ be two polynomials and $a$ be a real number. Then

1. Polynomial functions:

$lim_{x \to a} P (x) = P (a)$

2. Rational functions:

$lim_{x \to a} \frac{P (x)}{Q (x)} = \frac{P (a)}{Q (a)} (provided Q (a) \neq 0)$

Similar results hold for left and right hand limits; that is, we can replace $x \to a$ by $x \to a^{-}$ or $x \to a^{+}$ in the above equations.

Example 4

Evaluate $lim_{x \to - 0.5} (4 x^{2} - 3 x + 8)$ .

Solution

By Theorem 3:

$lim_{x \to - 0.5} (4 x^{2} - 3 x + 8) = 4 (- 0.5)^{2} - 3 (- 0.5) + 8 = 4 \times 0.25 + 1.5 + 8 = 10.5 .$

Example 5

Evaluate

$lim_{x \to - 2} \frac{3 x^{2} - x - 5}{x^{3} - 5 x}$

Solution

By Theorem 3:

$lim_{x \to - 2} \frac{3 x^{2} - x - 5}{x^{3} - 5 x} = \frac{3 (- 2)^{2} - (- 2) - 5}{(- 2)^{3} - 5 (- 2)} = \frac{9}{2} .$

Example 6

Let $g (t)$ be defined by

$g (t) = {\begin{cases} 4 + t & t > - 1 \\ 2 & t = - 1 \\ 2 t^{2} + 1 & t < - 1 \end{cases}$

Determine
(a) $lim_{t \to - 1^{-}} g (t)$
(b) $lim_{t \to - 1^{+}} g (t)$
(c) $lim_{t \to - 1} g (t)$

if they exist.

Solution

(a) For the left hand limit as $t \to - 1^{-}$ , we use the equation $g (t) = 2 t^{2} + 1$

$lim_{t \to - 1^{-}} g (t) = lim_{t \to - 1^{-}} (2 t^{2} + 1) = 2 (- 1)^{2} + 1 = 3.$

(b) For the right hand limit as $t \to - 1^{+ 1}$ , we use the equation $g (t) = 4 + t$

$lim_{t \to - 1^{+}} g (t) = lim_{t \to - 1^{+}} (4 + t) = 4 + (- 1) = 3.$

(c) Because $lim_{t \to - 1^{-}} g (t) = lim_{t \to - 1^{+}} g (t) = 3$ , by Theorem 1 in Section 4.3,

$lim_{t \to - 1} g (t) = 3.$

The graph of $y = g (t)$ is shown in the following figure.

Graph of

y = g (t)

Root Rule

In an example in Section 4.2, we showed that

$lim_{x \to a} \sqrt{x} = \sqrt{a}, (a > 0)$

Similarly, using the $ϵ$ - $δ$ definition, we can show that the following theorem is true. This limit is consistent with the appearance of the graph $y = \sqrt[n]{x}$ .

Theorem 4: If $n$ is a positive integer, then

$lim_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}$

[If $n$ is even, we assume that $a > 0$ .]

In general, we have the following theorem which will be shown in the Section on Continuity that it is a consequence of the above theorem.

Theorem 5 (Root Rule): If $n$ is a positive integer, then

$lim_{x \to a} \sqrt[n]{f (x)} = \sqrt[n]{lim_{x \to a} f (x)}$

[If $n$ is even we assume that $lim_{x \to a} f (x) > 0$ ]

To show that the above theorem is plausible, assume that $lim_{x \to a} \sqrt[n]{f (x)} = L$ exists. Then by the general form of the Product Rule,

$\begin{aligned} \underset{n times}{\underset{⏟}{(lim_{x \to a} \sqrt[n]{f (x)}) \dots (lim_{x \to a} \sqrt[n]{f (x)})}} & = lim_{x \to a} (\underset{n times}{\underset{⏟}{\sqrt[n]{f (x)} \dots \sqrt[n]{f (x)}}}) \\ \underset{n times}{\underset{⏟}{L \dots L}} & = lim_{x \to a} (f (x)) \end{aligned}$

$\Rightarrow L = \sqrt[n]{lim_{x \to a} f (x)} .$

Example 7

Evaluate the following limit and indicate the limit theorem being used for each step.

$lim_{x \to 2} \sqrt{\frac{x^{3} + 3 x + 2}{x^{2} + 5}}$

Solution

$\begin{aligned} lim_{x \to 2} \sqrt{\frac{x^{3} + 3 x + 2}{x^{2} + 5}} = & \sqrt{lim_{x \to 2} \frac{x^{3} + 3 x + 2}{x^{2} + 5}} & (by Thm 5) \\ = & \sqrt{\frac{2^{3} + 3 \times 2 + 2}{2^{2} + 5}} & (by Thm 3) \\ = & \sqrt{\frac{16}{9}} \\ = & \frac{4}{3} . \end{aligned}$

Replacement Rule

Suppose we want to evaluate

$lim_{x \to 1} \frac{x^{2} - 1}{x - 1}$

Because the denominator $x - 1$ is zero at $x = 1$ , we cannot use the above theorem. So what can we do here? Let\textquoteright s simplify this fraction

$\frac{x^{2} - 1}{x - 1} = \frac{(x - 1) (x + 1)}{x - 1} = x + 1 (if x \neq 1)$

This shows us that two functions $y = \frac{x^{2} - 1}{x - 1}$ and $y = x + 1$ are identical except when $x = 1$ . But the limit of $\frac{x^{2} - 1}{x - 1}$ depends only on the values of this function for $x$ near $1$ and the value of a function at $x = 1$ does not influence the limit. So we must have

$lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = lim_{x \to 1} x + 1$

By Theorem 3:

$lim_{x \to 1} (x + 1) = 1 + 1 = 2$

Therefore:

$lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = 2.$

In general recall that $lim_{x \to a} f (x)$ depends only on the values of $f (x)$ for $x$ close to $a$ and not on the value of $f$ at $x = a$ nor on the values of $f (x)$ for $x$ far away from $a$ . So if there is a function $g$ such that $f (x) = g (x)$ for $x$ close to $a$ but not necessarily for $x = a$ , then

$lim_{x \to a} f (x) = lim_{x \to a} g (x)$

Therefore, we have the following theorem:

Theorem 6 (Replacement Rule): If

$f (x) = g (x)$

for all $x$ close to $a$ , but not necessarily including $x = a$ , then

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

To be more precise, we can say if $f (x) = g (x)$ in a deleted neighborhood of $a$ then $lim_{x \to a} f (x) = lim_{x \to a} g (x)$ .

Illustration of this theorem is presented in the following figure.

We can see that $f (x) = g (x)$ for all $x$ between $b$ and $c$ (or in an open interval containing $a$ ) except for $x = a$ . Therefore $lim_{x \to a} f (x) = lim_{x \to a} g (x) = L$ .

Example 8

Given that $f$ is a function defined by

$f (t) = {\begin{cases} t + 3 & if t \neq - 1 \\ - 3 & if t = - 1 \end{cases},$

evaluate $lim_{t \to - 1} f (t)$ .

Solution

As discussed above, when evaluating the limit, we are concerned about the values of the variable (here $t$ ) near $- 1$ but not equal to $- 1$ . Thus we have,

$lim_{t \to - 1} f (t) = lim_{t \to - 1} (t + 3) = (- 1) + 3 = 2.$

Example 9

Evaluate

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

Solution

Here we cannot simply use Limits of Rational Functions (Theorem 3) and substitute 2 for $x$ in the given expression because $lim_{x \to 2} (3 x - 6) = 0$ . However, factoring the numerator and the denominator, we obtain

$\frac{x^{2} - 2^{2}}{3 x - 6} = \frac{(x - 2) (x + 2)}{3 (x - 2)} = {\begin{cases} \frac{1}{3} (x + 2) & if x \neq 2 \\ undefined & if x = 2 \end{cases}$

This quotient is $\frac{1}{3} (x + 2)$ when $x \neq 2$ . So by Replacement Rule (Theorem 6),

$\begin{aligned} lim_{x \to 2} \frac{x^{2} - 4}{3 x - 6} & = lim_{x \to 2} (\frac{1}{3} (x + 2)) \\ = \frac{1}{3} lim_{x \to 2} (x + 2) \\ = \frac{1}{3} (2 + 2) = \frac{4}{3} . \end{aligned}$

Example 10

Evaluate

$lim_{x \to - 3} \frac{x^{3} + 27}{x + 3}$

Solution

The substitution of $x = - 3$ in the fraction produces the meaningless fraction $0 / 0$ . So we cannot apply Limits of Rational Functions (Theorem 3). However, factoring the numerator, we obtain

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

[for factoring $x^{3} + 3^{3}$ recall a cube of SOAP (same, opposite,always positive) $A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})$ in Section on Factorization]

Because $\frac{x^{3} + 27}{x + 3}$ and $x^{2} - 3 x + 9$ are equal except when $x = - 3$ , by Theorem $???$

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

The Sandwich Theorem

The following theorem helps us find a variety of limits:

Theorem 7 (The Sandwich Theorem): If we have

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

for all $x$ in some open interval containing $a$ except possibly at $x = a$ itself and if

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

then we also have

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

The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.
The Sandwich Theorem also holds for the one-sided limits; that is, we can replace $x \to a$ by $x \to a^{-}$ or $x \to a^{+}$ in the above equations, and the theorem will be still valid.

The Sandwich Theorem states that if a function $f$ is sandwiched between two functions $g$ and $h$ near $a$ and if $g$ and $h$ approach the same number $L$ , then $f$ must approach $L$ too (as $x \to a$ , where else can $f$ go to, if not to $L$ ?). The validity of this theorem is suggested by the following figure.

The Sandwich Theorem. If

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

for all

x

near

a

(except possibly at

a

) and

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

, then

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

The rigorous proof is as follows.

Read the Proof of the Sandwich Theorem

Suppose $ϵ > 0$ is given. We need to show that there exists a $δ > 0$ such that for all $x$

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

Because $lim_{x \to a} g (x) = L$ and $lim_{x \to a} h (x) = L$ , there exist some $δ_{1} > 0$ and $δ_{2} > 0$ such that for all $x$

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

and

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

[Note that $| g (x) - L | < ϵ$ is equivalent to $- ϵ < g (x) - L < ϵ$ or if we add $L$ to each side it is equivalent to $L - ϵ < g (x) < L + ϵ$ . See Property 7(i) in the Section on Absolute Value]

Now if we choose $δ = min {δ_{1}, δ_{2}}$ , then for all $x$ if $0 < | x - a | < δ$ then

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

That is,

$L - ϵ < f (x) < L + ϵ$

$| f (x) - L | < ϵ .$

Example 11

Show that

$lim_{x \to 0} x ⌊ \frac{1}{x} ⌋ = 1,$

where the symbol $⌊ ⌋$ (also $[[]]$ ) denotes the greatest integer (or floor) function.

Solution

For every number $t$ , we know

$t - 1 < ⌊ t ⌋ \leq t .$

Therefore, for every $x \neq 0$

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

If $x > 0$ , we can multiply each side by $x$ and the directions of inequalities are preserved:

$1 - x < x ⌊ \frac{1}{x} ⌋ \leq 1.$

Because $lim_{x \to 0} (1 - x) = 1 - 0 = 1$ , by the Sandwich Theorem:

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

If $x < 0$ , we still multiply both sides by $x$ but the directions of inequalities are reversed:

$1 - x > x ⌊ \frac{1}{x} ⌋ \geq 1.$

Again because $lim_{x \to 0^{-}} (1 - x) = 1 - 0 = 1$ , by the Sandwich Theorem

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

Because $lim_{x \to 0^{-}} x ⌊ \frac{1}{x} ⌋ = lim_{x \to 0^{+}} x ⌊ \frac{1}{x} ⌋ = 1$ , we conclude that

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

The graphs of $y = x ⌊ \frac{1}{x} ⌋$ , $y = 1 - x$ , and $y = 1$ are depicted in the following figure.

Graphs of

y = x ⌊ \frac{1}{x} ⌋

y = 1 - x

, and

y = 1

Trigonometric Functions

In the Section on the Trigonometric Inequalities, we learned that for all values of $x$ , we have

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

and

$- | x | \leq 1 - \cos x \leq | x | .$

We can use the above inequalities and the Sandwich Theorem to solve the next example.

Example 12

Use the sandwich theorem to prove

(a) $lim_{x \to 0} \sin x = 0$

(b) $lim_{x \to 0} \cos x = 1$ .

Solution

Because

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

by the sandwich theorem, we have

(a)

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

(b)

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

By the Difference Rule (in Theorem 2), we have

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

Because $lim_{x \to 0} 1 = 1$ , we obtain

$1 - lim_{x \to 0} \cos x = 0 \Rightarrow lim_{x \to 0} \cos x = 1.$

Example 13

Use the previous example, to prove

(a) $lim_{x \to a} \sin x = \sin a$

(b) $lim_{x \to a} \cos x = \cos a$ .

Solution

Let

u = x - a

. In this case,

x \to a

is equivalent to

u \to 0

and

$lim_{x \to a} \sin x = lim_{u \to 0} \sin (u + a), lim_{x \to a} \cos x = lim_{u \to 0} \cos (u + a) .$

(a) Using the addition formula for sine, we have (see Equation 6 in the Section on Trignometric Identities)

$\sin (u + a) = \sin u \cos a + \cos u \sin a$

Thus

$lim_{u \to 0} \sin (u + a) = lim_{u \to 0} (\cos a \sin u + \sin a \cos u) .$

Because $\sin a$ and $\cos a$ are two constants, by the Constant Multiple Rule (Theorem 2),

we have

$lim_{u \to 0} (\cos a \sin u + \sin a \cos u) = \cos a lim_{u \to 0} \sin u + \sin a lim_{u \to 0} \cos u .$

In the previous exmple, we learned $lim_{u \to 0} \sin u = 0$ and $lim_{u \to 0} \cos u = 1$ . Therefore

$\cos a lim_{u \to 0} \sin u + \sin a lim_{u \to 0} \cos u = \cos a \times 0 + \sin a \times 1 = \sin a$

Finally

$lim_{x \to a} \sin x = lim_{u \to 0} \sin (u + a) = \sin a$

(b) Similar to part (a), we use the addition formula but for cosine (see Equation 7 in the Section on Trignometric Identities)

$\begin{aligned} lim_{u \to 0} \cos (u + a) & = lim_{u \to 0} (\cos u \cos a + \sin u \sin a) \\ = \cos a lim_{u \to 0} \cos u + \sin a lim_{u \to 0} \sin u \\ = \cos a \times 1 + \sin a \times 0 \\ = \cos a . \end{aligned}$

Example 14

Evaluate the following limits

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

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

Solution

(a) Note that for $x \neq 0$ , we have

$\begin{matrix} (i) & - 1 \leq \sin \frac{1}{x} \leq 1. \end{matrix}$

If $x > 0$ and we multiply each side of (i) by $x$ , the directions of the inequalities will not change. That is,

$- x \leq x \sin \frac{1}{x} \leq x . (x > 0)$

Because

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

it follows from the Sandwich Theorem that

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

If $x < 0$ and we multiply each side of (i) by $x$ , we need to reverse the direction of the inequalities. That is,

$- x \geq x \sin \frac{1}{x} \geq x . (x < 0)$

Again because

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

it follows from the Sandwitch Theorem that

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

Because the left and right limits are equal, we conclude that

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

The graph of $y = x \sin (1 / x)$ is shown below

(b) Similar to part (a), we start from the fact that

$- 1 \leq \cos t \leq 1$

and if we replace $t$ with $1 / x^{2}$ when $x \neq 0$ , we get

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

Multiplying each side by $\sqrt{x}$

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

Because

$lim_{x \to 0^{+}} \sqrt{x} = lim_{x \to 0^{+}} (- \sqrt{x}) = 0$

by the Sandwitch theorem

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

The graph of $y = \sqrt{x} \cos (1 / x^{2})$ is shown below.

(c) Similar to part (a), we start with

$- 1 \leq \sin (\frac{1}{x}) \leq 1 (if x \neq 0) .$

If $0 < x < π$ , then $\sin x > 0$ and therefore

$- \sin x \leq \sin x \sin (\frac{1}{x}) \leq \sin x (0 < x < π)$

and if $- π < x < 0$ , then $\sin x < 0$ and therefore

$- \sin x \geq \sin x \sin (\frac{1}{x}) \geq \sin x . (- π < x < 0)$

Because

$lim_{x \to 0} \sin x = lim_{x \to 0} (- \sin x) = 0$

it follows from the Sandwich Theorem that

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

The graph of $y = \sin x \cdot \sin (1 / x)$ is shown below.

Limits Involving sin(x)/x

Recall that when there is no $^{\circ}$ notation, the default is that $x$ is measured in radian.

In Example 3 in the Section on the Concept of a Limit [you need to click on “Show Some Examples” to be able to see this example], we saw that $lim_{x \to 0} \sin x / x = 1$ . This limit is of importance and we can solve many similar exercises using this limit. In this section, we prove that $lim_{x \to 0} \sin x / x = 1$ using the Sandwich Theorem.

Theorem 8:
$lim_{x \to 0} \frac{\sin x}{x} = 1.$

To prove this theorem, we show that the right-hand and left-hand limits are both 1, and consequently, the limit is 1.

Show the Proof

Consider the construction shown in the following figure, where the radius of the circle is 1 ( $O P = O A = 1$ ), and $0 < x < π / 2$ .
Notice that
$area of △ O A P < area of sector O A P < area of △ O A T$

△ O P H

\sin x = \frac{P H}{O P} = \frac{P H}{1} = P H

. In

△ O A T

\tan x = \frac{A T}{O A} = \frac{A T}{1} = A T

Let’s find each area in terms of $x$ :
$\begin{aligned} area of △ O A P & = \frac{1}{2} base \times height \\ = \frac{1}{2} O A \times P H \\ = \frac{1}{2} (1) (\sin x) \\ = \frac{1}{2} \sin x . \end{aligned}$
Because the central angle of the sector is $x$ :
$\begin{aligned} area of sector O A P & = \frac{1}{2} r^{2} θ \\ = \frac{1}{2} (1) x \\ = \frac{x}{2} . \end{aligned}$
And
$\begin{aligned} area of △ O A T & = \frac{1}{2} base \times height \\ = \frac{1}{2} O A \times A T \\ = \frac{1}{2} (1) (\tan x) \\ = \frac{1}{2} \tan x . \end{aligned}$
Thus,

$area of △ O A P < area of sector O A P < area of △ O A T$ $\Rightarrow \frac{1}{2} \sin x < \frac{1}{2} x < \frac{1}{2} \tan x$ or
$\sin x < x < \frac{\sin x}{\cos x} .$

Because $0 < x < π / 2$ , the sine of $x$ is positive and dividing each term by $\sin x$ does not change the directions of inequalities:
$1 < \frac{x}{\sin x} < \frac{1}{\cos x} .$ Taking reciprocals reverses the inequalities (see property 7 of inequalities in the Section on Inequalities):
$1 > \frac{\sin x}{x} > \cos x .$ Now we show that the last double inequality holds also for $- π / 2 < x < 0$ . If $- π / 2 < x < 0$ , then $0 < - x < π / 2$ and therefore
$1 > \frac{\sin (- x)}{(- x)} > \cos (- x) .$ Recall that $\sin x$ is an odd function and $\cos x$ is an even function. That is,
$\sin (- x) = - \sin x, \cos (- x) = \cos x$ Thus
$1 > \frac{- \sin x}{- x} > \cos (- x) = \cos x$ So we proved
$1 < \frac{\sin x}{x} < \cos x . (- \frac{π}{2} < x < \frac{π}{2}, x \neq 0)$ Because $lim_{x \to 0} \cos x = \cos 0 = 1$ and $lim_{x \to 0} 1 = 1$ , the Sandwich Theorem gives
$lim_{x \to 0} \frac{\sin x}{x} = 1.$

Note that
$lim_{x \to 0} \frac{x}{\sin x} = 1$ because
$\begin{aligned} lim_{x \to 0} \frac{x}{\sin x} & = lim_{x \to 0} \frac{1}{\frac{\sin x}{x}} \\ = \frac{lim_{x \to 0} 1}{lim_{x \to 0} \frac{\sin x}{x}} \\ = \frac{1}{1} = 1 \end{aligned}$

Example 15

Evaluate the following limit

$lim_{x \to 0} \frac{\sin (\sqrt{2} x)}{x}$

Solution

Let’s multiply the numerator and the denominator by $\sqrt{2}$

$lim_{x \to 0} \frac{\sin \sqrt{2} x}{x} = lim_{x \to 0} \frac{\sqrt{2} \sin (\sqrt{2} x)}{\sqrt{2} x}$

Now let $u = \sqrt{2} x$ . So $x \to 0$ is equivalent to $u \to 0$

$\begin{aligned} lim_{x \to 0} \frac{\sin \sqrt{2} x}{x} & = lim_{x \to 0} \frac{\sqrt{2} \sin (\sqrt{2} x)}{\sqrt{2} x} \\ = lim_{u \to 0} \sqrt{2} \frac{\sin u}{u} \\ = \sqrt{2} lim_{u \to 0} \frac{\sin u}{u} \\ = \sqrt{2} \times 1 = \sqrt{2} . \end{aligned}$

In general for a constant $A$

$lim_{x \to 0} \frac{\sin A x}{x} = A .$

Example 16

Show that for nonzero constants $A$ and $B$

$lim_{x \to 0} \frac{\sin A x}{\sin B x} = \frac{A}{B}$

Solution

Let’s divide both the denominator and numerator by $x$

$\begin{aligned} lim_{x \to 0} \frac{\sin A x}{\sin B x} & = lim_{x \to 0} \frac{\frac{\sin A x}{x}}{\frac{\sin B x}{x}} \\ = \frac{lim_{x \to 0} \frac{\sin A x}{x}}{lim_{x \to 0} \frac{\sin B x}{x}} \end{aligned}$

We just saw that $lim_{x \to 0} \frac{\sin A x}{x} = A$ . Thus,

$\begin{aligned} lim_{x \to 0} \frac{\sin A x}{\sin B x} & = \frac{lim_{x \to 0} \frac{\sin A x}{x}}{lim_{x \to 0} \frac{\sin B x}{x}} \\ = \frac{A}{B} . \end{aligned}$

Example 17

Show that

$lim_{x \to 0} \frac{\tan x}{x} = 1$

Solution

By definition $\tan x = \cos x / \sin x$ . Thus

$\begin{aligned} lim_{x \to 0} \frac{\tan x}{x} & = lim_{x \to 0} \frac{\frac{\sin x}{\cos x}}{x} \\ = lim_{x \to 0} \frac{1}{\cos x} \frac{\sin x}{x} \\ = lim_{x \to 0} \frac{1}{\cos x} \cdot lim_{x \to 0} \frac{\sin x}{x} \\ = \frac{1}{\cos 0} \cdot 1 \\ = (1) (1) \\ = 1. \end{aligned}$
\end{sol}

$lim_{x \to 0} \frac{\tan x}{x} = 1$

Example 18

Evaluate the following limit

$lim_{x \to 0} \frac{\sin (\sin x)}{x}$

Solution

Let’s multiply and divide the given fraction by $\sin x$ . That is,

$\begin{aligned} lim_{x \to 0} \frac{\sin (\sin x)}{x} & = lim_{x \to 0} \frac{\sin (\sin x)}{\sin x} \frac{\sin x}{x} \\ = lim_{x \to 0} \frac{\sin (\sin x)}{\sin x} \cdot lim_{x \to 0} \frac{\sin x}{x} \end{aligned}$

For the first limit let $u = \sin x$ . If $x \to 0$ , then $u = \sin x \to 0$ . Thus

$\begin{aligned} lim_{x \to 0} \frac{\sin (\sin x)}{x} & = lim_{x \to 0} \frac{\sin (\sin x)}{\sin x} \cdot lim_{x \to 0} \frac{\sin x}{x} \\ = lim_{u \to 0} \frac{\sin u}{u} \cdot lim_{x \to 0} \frac{\sin x}{x} \\ = 1 \cdot 1 \\ = 1. \end{aligned}$

Another importance limit is

$lim_{x \to 0} \frac{1 - \cos x}{x^{2}}$

Notice that we cannot simply substitute $x = 0$ in the given expression because we get the meaningless fraction $0 / 0$ upon substitution.

Recall that the half-angle formula:

$\frac{1 - \cos 2 θ}{2} = \sin^{2} θ$

Now let $x = 2 θ$ . Therefore,

$1 - \cos x = 2 \sin^{2} (\frac{x}{2})$

and

$\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} & = lim_{x \to 0} \frac{2 \sin^{2} (\frac{x}{2})}{x^{2}} \\ = 2 lim_{x \to 0} \frac{\sin (x / 2)}{x} \cdot \frac{\sin (x / 2)}{x} \\ = 2 lim_{x \to 0} \frac{\sin (x / 2)}{x} \cdot lim_{x \to 0} \frac{\sin (x / 2)}{x} \end{aligned}$

We just saw that if $A$ is a constant

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

Therefore

$lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = 2 (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{2} .$

$lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2} .$

From the above limit, we can conclude that

$lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

because

$\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{x} & = lim_{x \to 0} (x \frac{1 - \cos x}{x^{2}}) \\ = (lim_{x \to 0} x) (lim_{x \to 0} \frac{1 - \cos x}{x^{2}}) \\ = (0) (\frac{1}{2}) = 0. \end{aligned}$

Example 19

Find $lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}}$ .

Solution

We cannot simply substitute $x = 0$ in the given fraction because we will get the meaningless fraction $0 / 0$ . Instead we use the definition of $\tan x = \sin x / \cos x$

$\begin{aligned} lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}} & = lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^{3}} \\ = lim_{x \to 0} \frac{\frac{\sin x - \sin x \cos x}{\cos x}}{x^{3}} \\ = lim_{x \to 0} \frac{\frac{\sin x (1 - \cos x)}{\cos x}}{x^{3}} \\ = lim_{x \to 0} (\frac{1}{\cos x} \frac{\sin x (1 - \cos x)}{x^{3}}) \\ = lim_{x \to 0} (\frac{1}{\cos x} \frac{\sin x}{x} \frac{1 - \cos x}{x^{2}}) \\ = (lim_{x \to 0} \frac{1}{\cos x}) (lim_{x \to 0} \frac{\sin x}{x}) (lim_{x \to 0} \frac{1 - \cos x}{x^{2}}) \\ = (\frac{1}{\cos 0}) (1) (\frac{1}{2}) \\ = \frac{1}{2} . \end{aligned}$

Example 20

Show that

$lim_{x \to 0} \frac{x^{2} \sin (\frac{1}{x})}{\sin x} = 0$

Solution

We use the following two limits that we already know

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

So we can write

$\begin{aligned} lim_{x \to 0} \frac{x^{2} \sin \frac{1}{x}}{\sin x} & = lim_{x \to 0} (\frac{x}{\sin x} x \sin \frac{1}{x}) \\ = (lim_{x \to 0} \frac{x}{\sin x}) (lim_{x \to 0} x \sin \frac{1}{x}) \\ = (lim_{x \to 0} \frac{1}{\frac{\sin x}{x}}) (lim_{x \to 0} x \sin \frac{1}{x}) \\ = (\frac{lim_{x \to 0} 1}{lim_{x \to 0} \frac{\sin x}{x}}) (lim_{x \to 0} x \sin \frac{1}{x}) \\ = (\frac{1}{1}) (0) \\ = 0 \end{aligned}$

Example 21

Find

$lim_{x \to 0} \frac{\arcsin x}{x}$

[In some books the inverse of sine is denoted by $\sin^{- 1} x$ instead of $\arcsin x$ ]

Solution

Let

$u = \arcsin x \Rightarrow \sin u = \sin (\arcsin x) = x$

[Recall $\sin (\arcsin x) = x$ for $- 1 \leq x \leq 1.$ ]

We know as $x \to 0$ , $u = \arcsin x \to 0$ . Therefore

$\begin{aligned} lim_{x \to 0} \frac{\arcsin x}{x} & = lim_{u \to 0} \frac{u}{\sin u} = lim_{u \to 0} \frac{1}{\frac{\sin u}{u}} \\ = \frac{lim_{u \to 0} 1}{lim_{u \to 0} \frac{\sin u}{u}} \\ = \frac{1}{1} = 1. \end{aligned}$

Algebraic Operations on Limits

Show the Proof of Theorem 1

Hide the Proof of Theorem 1

Basic Algebraic Operations

Limits of Polynomial and Rational Functions

Root Rule

Replacement Rule

The Sandwich Theorem

Read the Proof of the Sandwich Theorem

Hide the Proof

Trigonometric Functions

Limits Involving sin(x)/x

Show the Proof

Hide the Proof