$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin 3 x \cot 4 x } { \cot 5 x }$$
A) 0
B) $\frac { 15 } { 4 }$
C) does not exist
D) $\frac { 12 } { 5 }$

Question

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The Answer of  $\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin 3 x \cot 4 x } { \cot 5 x }$$
A) 0
B) $\frac { 15 } { 4 }$
C) does not exist
D) $\frac { 12 } { 5 }$

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.

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