Question

$\underset{\theta \to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}-\cos \theta -\sin \theta }{(4\theta -\pi {)}^2}$ 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)

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