Evaluate the following limits:


Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{\sin 2 x(1-\cos 2 x)}{x^{3}}$



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, $\lim _{x \rightarrow 0} \frac{\sin 2 x(1-\cos 2 x)}{x^{3}}=4$


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