_ x 2 1+ 2 x ( -2 x)^2 = - Vidyadip

MCQ

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

A
1
B
2
C
3
✓
$\frac{1}{2}$

✓

Answer

Correct option: D.

$\frac{1}{2}$

(D)
Put $\pi-2 x=\theta$
$\Rightarrow 2 x=\pi-\theta$ and as $x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}=\lim _{\theta \rightarrow 0} \frac{1+\cos (\pi-\theta)}{\theta^2}$
$=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2}=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 \frac{\theta}{2}}{\frac{\theta^2}{4} \times 4}=\frac{1}{2} \lim _{\theta \rightarrow 0}\left(\frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}\right)^2$
$=\frac{1}{2}( l )^2=\frac{1}{2}$

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