$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin 5 x } { x }$$
A) does not exist
B) 1
C) $\frac { 1 } { 5 }$
D) 5

Question

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin 5 x } { x }$$ 
A) does not exist
B) 1
C) $\frac { 1 } { 5 }$
D) 5

Quizplus · Accepted Answer

Using the given limit $\lim _ { x = 0 } \frac { \sin x } { x } = 1$, we can transform the expression $\frac { \sin 5x } { x }$ to $\frac { 5 \sin 5x } { 5x }$ which equals $5 \cdot \lim _ { x = 0 } \frac { \sin 5x } { 5x } = 5\cdot 1 = 5$, since $\lim _ { x = 0 } \frac { \sin 5x } { 5x } = 1$ by the given property.

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.

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