# Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow \pi} \frac{\sin 3 x-3 \sin x}{(\pi-x)^{3}}$

Solution:

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is $\frac{0}{0}$

By using $L$ hospital Rule,

Differtiate both sides w.r.t $x$

So $\lim _{x \rightarrow \pi} \frac{\sin 3 x-3 \sin x}{(\pi-x)^{3}}=\lim _{x \rightarrow \pi} \frac{3 \cos 3 x-3 \cos x}{-3(\pi-x)^{2}}$

Again, indeterminate Form is $\frac{0}{0}$

So, Differtiate both sides w.r.t $x$ again, we have

$\lim _{x \rightarrow \pi} \frac{\sin 3 x-3 \sin x}{(\pi-x)^{3}}=\lim _{x \rightarrow \pi} \frac{-9 \sin 3 x+3 \sin x}{6(\pi-x)}$

Again, indeterminate Form is $\frac{0}{0}$

So, Differtiate both sides w.r.t $x$ again, we have

$\lim _{x \rightarrow \pi} \frac{\sin 3 x-3 \sin x}{(\pi-x)^{3}}=\lim _{x \rightarrow \pi} \frac{-27 \cos 3 x+3 \cos x}{-6}=\frac{-27 \cos 3 \pi+3 \cos \pi}{-6}=\frac{27-3}{-6}=-4$

Therefore, $\lim _{x \rightarrow \pi} \frac{\sin 3 x-3 \sin x}{(\pi-x)^{3}}=-4$