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 are $\infty \times \infty$

$\operatorname{cosec} 2 x-\cot 2 x=(1-\cos 2 x) / \sin 2 x$

$\lim _{x \rightarrow 0} \frac{\cot 2 x-\operatorname{cosec} 2 x}{x}=\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{x \sin 2 x}=\lim _{x \rightarrow 0} \frac{\frac{\cos 2 x-1}{x^{2}}}{\frac{x \sin 2 x}{x^{2}}}=\lim _{x \rightarrow 0} \frac{\frac{4[\cos 2 x-1]}{(2 x)^{2}}}{\frac{2 \sin 2 x}{2 x}}$

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

$\lim _{x \rightarrow 0} \frac{\cot 2 x-\operatorname{cosec} 2 x}{x}=\lim _{x \rightarrow 0} \frac{\frac{4[\cos 2 x-1]}{(2 x)^{2}}}{\frac{2 \sin 2 x}{2 x}}=\frac{-4}{2}=-2$

Therefore, $\lim _{x \rightarrow 0} \frac{\cot 2 x-\operatorname{cosec} 2 x}{x}=-2$