Question

$\displaystyle\lim _{\theta \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}-\cos \,\theta-\sin \,\theta}{(4 \,\theta-\pi)^{2}}$ is equal to

• A. $\frac{1}{16}$
• B. $\frac{1}{16 \sqrt{2}}$
• C. $\frac{1}{8}$
• D. $\frac{1}{8 \sqrt{2}}$
• E. $\frac{1}{10}$

Answer 1

Answer: (B)

$\displaystyle\lim _{0 \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}-\sqrt{2} \cos \left(\theta-\frac{\pi}{4}\right)}{16\left(\theta-\frac{\pi}{4}\right)^{2}}$
$=\displaystyle\lim _{y \rightarrow 0} \frac{\sqrt{2}}{16} \cdot \frac{( l -\cos y)}{y^{2}}$
Where, $y=\theta-\frac{\pi}{4} \rightarrow 0$ as $\theta=\frac{\pi}{4}$
$=\frac{1}{8 \sqrt{2}} \displaystyle\lim _{y \rightarrow 0} \frac{2 \sin ^{2}\left(\frac{y}{2}\right)}{y^{2}}$
$=\frac{1}{8 \sqrt{2}} \cdot \frac{1}{2}=\frac{1}{16 \sqrt{2}}$