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

Question

Quizplus · Accepted Answer

The given limit can be simplified using the identity $\lim _ { x = 0 } \frac { \sin x } { x } = 1$. We rewrite the expression as $\frac{\sin x}{x} \cdot \frac{\cos 4x}{1 + \cos 5x}$. As $x \rightarrow 0$, $\frac{\sin x}{x} \rightarrow 1$ and $\cos 4x \rightarrow \cos 0 = 1$, $\cos 5x \rightarrow \cos 0 = 1$. Thus, 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.

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