Chapter 14.2, Problem 53AYU

In problems 53-56, use the properties of limits and the facts that $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0\underset{x\to 0}{\lim }\sin x=0\underset{x\to 0}{\lim }\cos x=1$ where $x$ is in radians, to find each limit.

$\underset{x\to 0}{\lim }\frac{\tan x}{x}$

To determine

To find: The limit ${\lim }_{x\to \text{ 0}}\frac{\tan x}{x}$, use the properties of limits and facts that ${\lim }_{x\to \text{ 0}}\frac{\sin x}{x}=1$; ${\lim }_{x\to \text{ 0}}\frac{\text{cos}x - 1}{x}=0$; ${\lim }_{x\to \text{ 0}}\sin x=0$; ${\lim }_{x\to \text{ 0}}\sin x=0=1$.

Answer to Problem 53AYU

Solution:

1

Explanation of Solution

Given:

${\lim }_{x\to \text{ 0}}\frac{\tan x}{x}$

Formula used:

${\lim }_{x\to \text{ c}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to c}{\text{lim}}f\left(x\right)}{\underset{x\to c}{\text{lim}}g\left(x\right)}$, provided that ${\lim }_{x\to \text{ 1}}g\left(x\right)\ne 0$

Calculation:

For, ${\lim }_{x\to \text{ 0}}\frac{\tan x}{x}$, at $f\left(0\right)$ the limit of the denominator is zero, which is undefined,

so we have to simplify it,

$\frac{\tan x}{x}=\frac{\frac{\sin x}{\cos x}}{x}=\frac{\sin x}{x\left(\cos x\right)}=\frac{\frac{\sin x}{x}}{\cos x}$

Now the quotient formula can be used.

${\lim }_{x\to \text{ c}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to c}{ \text{lim}}f\left(x\right)}{\underset{x\to c}{ \text{lim}}g\left(x\right)}$, provided that $\underset{x\to c}{\text{lim}}g\left(x\right)\ne 0$

${\lim }_{x\to \text{ 0}}\left[\frac{\frac{\sin x}{x}}{\cos x}\right]=\frac{\underset{x\to 0}{\text{lim}\frac{\sin x}{x}}}{\underset{x\to 0}{\text{ lim}}\cos x}=\frac{1}{1}=1$