Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^{3}}$
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 $\frac{0}{0}$
$\lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sin 2 x+\sin 2 x \cos 2 x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sin 2 x(1-\cos 2 x)}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin 2 x}{2 x} \times \frac{4(1-\cos 2 x)}{(2 x)^{2}}=4$
Therefore, $\lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^{3}}=4$