Question
Evaluate $\lim _{x \rightarrow \pi} \frac{\sin ^2 2 x}{\cos ^2 \frac{1}{2} x}.$

Answer


$\begin{array}{l}\lim _{x \rightarrow \pi} \frac{\sin ^2 2 x}{\cos ^2 \frac{1}{2} x}=\lim _{x \rightarrow \pi} \frac{(2 \sin x \cos x)^2}{\cos ^2 \frac{x}{2}} \\\because \sin 2 x=2 \sin x \cos x\\ =\lim _{x \rightarrow \pi} \frac{4 \sin ^2 x \cos ^2 x}{\cos ^2 \frac{1}{2} x}=\lim _{x \rightarrow \pi} \frac{4(\sin x)^2 \cos ^2 x}{\cos ^2 \frac{x}{2}} \\ =\lim _{x \rightarrow \pi} \frac{4.4 \sin ^2 \frac{x}{2} \cdot \cos ^2 \frac{x}{2} \cdot \cos ^2 x}{\cos ^2 \frac{x}{2}} \because \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \\ =\lim _{x \rightarrow \pi}\left(16 \sin ^2 \frac{x}{2} \cdot \cos ^2 x\right) \\ =16 \sin ^2 \frac{\pi}{2} \cdot \cos ^2 \pi=16(1)^2(-1)^2 \\ =16.1 .1=16\end{array}$

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