$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin x \cos 4 x } { x + x \cos 5 x }$$
A) $\frac { 1 } { 2 }$
B) does not exist
C) $\frac { 4 } { 5 }$
D) 0

Question

Quizplus · Accepted Answer

The limit can be simplified using the given $\lim _ { x = 0 } \frac { \sin x } { x } = 1$. Simplify the expression to $\frac{\sin x \cos 4x}{x(1 + \cos 5x)}$. Since $\lim _ { x = 0 } \cos nx = 1$ for any n, the limit becomes $\frac{1 \cdot 1}{1 + 1} = \frac{1}{2}$.

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.

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