$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow0 } \frac { x } { \sin 3 x }$$
A) 3
B) 1
C) $\frac { 1 } { 3 }$
D) does not exist

Question

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow0 } \frac { x } { \sin 3 x }$$ 
A) 3
B) 1
C) $\frac { 1 } { 3 }$
D) does not exist

Quizplus · Accepted Answer

We can use the limit provided in the problem statement to rewrite the given limit as follows:

$$\lim_{x \rightarrow 0} \frac{x}{\sin 3x} = \lim_{x \rightarrow 0}\frac{1}{3}\cdot\frac{3x}{\sin 3x} = \frac{1}{3}\cdot \lim_{x \rightarrow 0}\frac{\sin 3x}{3x}$$

Using the given limit, we know that $\lim_{x \rightarrow 0}\frac{\sin 3x}{3x} = 1$, so:

$$\frac{1}{3}\cdot \lim_{x \rightarrow 0}\frac{\sin 3x}{3x} = \frac{1}{3}\cdot(1) = \boxed{\frac{1}{3}}$$

Therefore, the best choice is (C) with the given explanation.

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.

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