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