$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { x } { \sin 3 x }$$
A) 3
B) 1
C) does not exist
D) $\frac { 1 } { 3 }$

Question

Quizplus · Accepted Answer

We can use the fact that $\lim_{x	o 0}\frac{\sin x}{x}=1$ to manipulate the original limit:
\begin{align*}
\lim_{x	o 0}\frac{x}{\sin 3x} &= \lim_{x	o 0}\frac{1}{\frac{\sin 3x}{x}}\
&=\lim_{x	o 0}\frac{1}{\frac{\sin 3x}{3x}\cdot 3}\
&=\frac{1}{3}\left(\lim_{x	o 0}\frac{\sin 3x}{3x}ight)^{-1}\
&=\frac{1}{3},
\end{align*}
where we used the fact that $\lim_{x	o 0}\frac{\sin 3x}{3x}=1$ in the last step. Therefore, the answer is choice $	extbf{(D)}$.

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.

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