Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Laws of Limits

In this section, we combine all of our previous results and definitions to evaluate limits of functions that are the sum, product, quotient or composition of two or more functions. The next theorem gives the conditions for the arithmetic operations of limits that are known as “Laws of Limits.”

Theorem 3.23. Laws of Limits.

If $c$ is a constant and $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M$, then

1. $\underset{x\to a}{\lim }f\left(x\right)\pm g\left(x\right)=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)=L\pm M$
2. $\underset{x\to a}{\lim }cf\left(x\right)=c\underset{x\to a}{\lim }f\left(x\right)=cL$
3. $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }g\left(x\right)=LM$
4. ${\displaystyle \underset{x\to a}{\lim }{\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}={\displaystyle \frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}}={\displaystyle \frac{L}{M}}}$ if $M\ne 0$

This theorem holds if $x\to a$ is replaced by $x\to a^{+}$ or $x\to a^{-}\text{.}$

For the composition of functions, we have the following results.

Theorem 3.24. If $\underset{x\to a}{\lim }g\left(x\right)=b$ and $\underset{x\to b}{\lim }f\left(x\right)=L$, then

$\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=L\text{.}$

This theorem also holds if $x\to a$ is replaced by $x\to a^{+}$ or $x\to a^{-}\text{.}$

We can see that if $u=g\left(x\right)$, then $u=g\left(x\right)\to b$ as $x\to a$, and Theorem 3.24 states that

$\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=\underset{u\to b}{\lim }f\left(u\right)=L\text{.}$

Two particular cases of Theorem 3.24 are important. The first occurs when the function $f$ is continuous at $b\text{.}$ In this case, $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$ and we have the following corollary.

Corollary 3.25. If $\underset{x\to a}{\lim }g\left(x\right)=b$ and $f$ is continuous at $b$, then

$\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=f\left(b\right)\text{.}$

The second case occurs when the outside function $f$ is the power function $f\left(x\right)=x^r\text{.}$ Then the function $f$ is continuous at $b$ if $\underset{x\to b}{\lim }{x}^r=b^r$ (the number $b^r$ is well defined). Hence, a direct application of Theorem 3.24 gives the following corollary.

Corollary 3.26. If $\underset{x\to a}{\lim }g\left(x\right)=b$ and $f\left(x\right)=x^r$ is continuous at $b$, then

$\underset{x\to a}{\lim }f(g(x))=\underset{x\to a}{\lim }{(g(x))}^r={\left(\underset{x\to a}{\lim }g(x)\right)}^r=b^r$

This corollary holds if $x\to a$ is replaced by $x\to a^{+}$ or $x\to a^{-}\text{.}$

Observe that in order to apply Theorem 3.23, we must know that the limits of $f$ and $g$ are finite. To apply the Laws of Limits, we may need to manipulate the function first, and use Proposition 3.16.

Example 3.57. The function

${\displaystyle f\left(x\right)={\displaystyle \frac{x}{\sin x}}}$

resembles the function in the limit of Example 3.56, above. We change it so that we can use this limit.

We have

${\displaystyle f\left(x\right)={\displaystyle \frac{x}{\sin x}}={\displaystyle \frac{1}{{\displaystyle \frac{\sin x}{x}}}}}$.

Applying Theorem 3.23(d), we get

$\underset{x\to 0}{\lim }{\displaystyle \frac{x}{\sin x}}=\underset{x\to 0}{\lim }{\displaystyle \frac{1}{{\displaystyle \frac{\sin x}{x}}}}={\displaystyle \frac{\underset{x\to 0}{\lim }1}{\underset{x\to 0}{\lim }{\displaystyle \frac{\sin x}{x}}}}={\displaystyle \frac{1}{1}}=1.$

Example 3.58. Let us evaluate the limit

$\underset{\theta \to 0}{\lim }{\displaystyle {\displaystyle \frac{\sin (c\theta )}{\theta }}}$

where $c$ is a nonzero constant.

We can see that if $t=c\theta$, then $\theta ={\displaystyle \frac{t}{c}}$ and $t\to 0$ as $\theta \to 0\text{.}$

Hence, by Theorem 3.23(b),

$\underset{\theta \to 0}{\lim }{\displaystyle {\displaystyle \frac{\sin (c\theta )}{\theta }}}=\underset{t\to 0}{\lim }{\displaystyle {\displaystyle \frac{c\sin (t)}{t}}}=c\underset{t\to 0}{\lim }{\displaystyle {\displaystyle \frac{\sin (t)}{t}}}=c(1)=c\text{.}$

Example 3.59. Similarly, to evaluate the limit

$\underset{\theta \to 0}{\lim }{\displaystyle {\displaystyle \frac{\theta }{\sin (c\theta )}}}\text{,}$

where $c$ is a nonzero constant, we use Example 3.57 and Theorem 3.23(c), setting $t=c\theta \text{.}$

We evaluate as follows:

$\underset{\theta \to 0}{\lim }{\displaystyle \frac{\theta }{\sin (c\theta )}}=\underset{t\to 0}{\lim }{\displaystyle \frac{t}{c\sin (t)}}={\displaystyle \frac{1}{c}}\underset{t\to 0}{\lim }{\displaystyle \frac{t}{(\sin t)}}={\displaystyle \frac{1}{c}}(1)={\displaystyle \frac{1}{c}}$

Example 3.60. If $c$ and $b$ are two nonzero constants, we see from Examples 3.58 and 3.59, above, and Theorem 3.23(c) that

$\underset{x\to 0}{\lim }{\displaystyle \frac{\sin (cx)}{\sin (bx)}}=\underset{x\to 0}{\lim }{\displaystyle \frac{\sin (cx)}{x}}{\displaystyle \frac{x}{\sin (bx)}}=\underset{x\to 0}{\lim }{\displaystyle \frac{\sin (cx)}{x}}\underset{x\to 0}{\lim } {\displaystyle \frac{x}{\sin (bx)}}=c{\displaystyle \frac{1}{b}}={\displaystyle \frac{c}{b}}\text{.}$

Example 3.61. We will use Theorem 3.14, Proposition 3.16, and Theorem 3.23 to evaluate the limit

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\cos x-1}{x}}\text{.}}$

First, we multiply the numerator by its conjugate to obtain the product of two functions whose limits at $0$ are finite.

$\begin{array}{rl}\left({\displaystyle \frac{\cos x-1}{x}}\right)\left({\displaystyle \frac{\cos x+1}{\cos x+1}}\right)& ={\displaystyle \frac{{\cos }^2-1}{x(\cos x+1)}}\\ & ={\displaystyle \frac{-{(\sin x)}^2}{x(\cos x+1)}}\\ & ={\displaystyle \frac{\sin x}{x}\frac{-\sin x}{\cos x+1}}\text{.}\end{array}$

By Theorem 3.14,

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin x}{\cos x+1}}=0\text{,}}$

and by Example 3.56,

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin x}{x}}=1\text{.}}$

Hence, by Proposition 3.16, and by Theorem 3.23(c),

$\begin{array}{rl}\underset{x\to 0}{\lim }{\displaystyle \frac{\cos x-1}{x}}& =\underset{x\to 0}{\lim }\left({\displaystyle \frac{\sin x}{x}}\right)\left({\displaystyle \frac{-\sin x}{\cos x+1}}\right)\\ & =\underset{x\to 0}{\lim }{\displaystyle \frac{\sin x}{x}}\underset{x\to 0}{\lim }{\displaystyle \frac{-\sin x}{\cos x+1}}\\ & =(1)(0)\\ & =0.\end{array}$

Example 3.62.

${\displaystyle \begin{array}{ll}\underset{x\to -1}{\lim }{\displaystyle \frac{\text{sin }\left(x+1\right)}{x^2-3x-4}}\hfill & =\underset{x\to -1}{\lim }{\displaystyle \frac{\sin \left(x+1\right)}{\left(x+1\right)\left(x-4\right)}}\hfill \\ \hfill & =\underset{x\to -1}{\lim }\left({\displaystyle \frac{\sin \left(x+1\right)}{x+1}}\right)\left({\displaystyle \frac{1}{x-4}}\right)\text{.}\hfill \end{array}}$

1. ${\displaystyle \underset{x\to -1}{\lim }{\displaystyle \frac{1}{x-4}}={\displaystyle \frac{1}{-5}}}$, by Theorem 3.14.

2. If $t=x+1$, then $t\to 0$ as $x\to -1$, and by Corollary 3.25 and Example 3.56,

   ${\displaystyle \underset{x\to -1}{\lim }{\displaystyle \frac{\sin \left(x+1\right)}{x+1}}=\underset{t\to 0}{\lim }{\displaystyle \frac{\sin t}{t}}=1\text{.}}$

   By Theorem 3.23,

   ${\displaystyle \begin{array}{ll}\underset{x\to -1}{\lim }{\displaystyle \frac{\text{sin }\left(x+1\right)}{x^2-3x-4}}\hfill & =\underset{x\to -1}{\lim }{\displaystyle \frac{\sin \left(x+1\right)}{\left(x+1\right)}}{\displaystyle \frac{1}{x-4}}\hfill \\ \hfill & =\underset{x\to -1}{\lim }{\displaystyle \frac{\sin \left(x+1\right)}{x+1}}\underset{x\to -1}{\lim }{\displaystyle \frac{1}{x-4}}\hfill \\ \hfill & =\left(1\right){\displaystyle \frac{1}{-5}}=-{\displaystyle \frac{1}{5}}\hfill \end{array}}$

Tutorial 3: Evaluation of a trigonometric limit with algebraic manipulation.