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

We know that $\sin 2 x=2 \sin x \cos x$

Formula used: $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

So, by using the above formula, we have

$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin x[1-\cos x]}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin x}{x} \times \frac{[1-\cos x]}{x^{2}}=\frac{2}{2}=1$

Therefore, $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}=1$