$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow } \frac { \tan 4 x } { x }$$
A) $\frac { 1 } { 4 }$
B) 1
C) does not exist
D) 4

Question

Quizplus · Accepted Answer

We know that 
$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$
$$\lim_{x\rightarrow 0} \sin x = 0$$

We can rewrite the original limit as 
$$\lim_{x\rightarrow 0} \frac{\tan 4x \cdot 4}{4x}$$
Using the identity 
$$\tan 4x = \frac{2\tan 2x}{1-\tan^2 2x}$$
We can rewrite the limit as 
$$\lim_{x\rightarrow 0} \frac{2\tan 2x \cdot 2}{4x - 4x\tan^2 2x}$$
$$= \lim_{x\rightarrow 0} \frac{2\tan 2x}{2-2\tan^2 2x}\cdot \frac{2}{x}$$
Using the fact that 
$$\lim_{x\rightarrow 0} \frac{\tan x}{x} = 1$$
We can replace $\tan 2x$ with $2x$ in the denominator, since $2x\rightarrow 0$ as $x\rightarrow 0$. Thus, the limit simplifies to 
$$= \lim_{x\rightarrow 0} \frac{2\cdot 2}{2} = \boxed{\textbf{(D) } 4}$$

Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } Find the limit using limx=0​xsinx​=1.

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$