Question
Evaluate $\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}.$

Answer


$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}$
Let $\frac{\pi}{2}+h=x$
$\begin{array}{l}\because \text { If } x=\frac{\pi}{2}, \text { then } h \rightarrow 0 \\=\lim _{h \rightarrow 0} \frac{1+\cos \left\{2\left(\frac{1}{2} \pi+h\right)\right\}}{\left\{\pi-2\left(\frac{1}{2}\pi+h\right)\right\}^2}\left[\because \lim _{x \rightarrow a} f(x)=\lim _{h \rightarrow 0} f(a+h)\right] \\=\lim _{h \rightarrow 0} \frac{1+\cos (\pi+2 h)}{4 h^2}=\lim _{h \rightarrow 0} \frac{1-\cos 2 h}{4 h^2}\end{array}$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{2 \sin ^2 h}{4 h^2}=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \\ =\frac{1}{2} \cdot(1)^2=\frac{1}{2}\end{array}$

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