Evaluate the following limits:

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, inderterminate Form is $\frac{0}{0}$

Formula used: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

So $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-\frac{2 \sin 3 x}{x}+\frac{\sin 5 x}{x}\right)=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-\frac{2 \sin 3 x}{3 x} \times 3+\frac{5 \sin 5 x}{5 x}\right)$

By using the above formula, we have

$\therefore \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-\frac{2 \sin 3 x}{3 x} \times 3+\frac{5 \sin 5 x}{5 x}\right)=1-2 \times 3+5=0$

Therefore, $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}=0$