Question

In Problems 53-56, use the properties of limits and the facts that

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin x}{x}=1 \\ \underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0 \\ \underset{x\to 0}{\lim }\sin x=0 \\ \underset{x\to 0}{\lim }\cos x=1 \end{gathered}$$

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

Short answer

The limit equals$1$

Step-by-step solution

Given information

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

Evaluating the limit by breaking into sine and cosine components

Calculating, we get

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\tan x}{x}=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}·\frac{1}{\cos x}\right) \\ \underset{x\to 0}{\lim }\frac{\tan x}{x}=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)·\underset{x\to 0}{\lim }\left(\frac{1}{\cos x}\right) \\ \underset{x\to 0}{\lim }\frac{\tan x}{x}=1·\frac{1}{1} \\ \underset{x\to 0}{\lim }\frac{\tan x}{x}=1 \end{gathered}$$