Question

Solve $\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sqrt{2-\cos \theta -\sin \theta }}{{(4\theta -\pi )}^2}$ is $\frac{1}{b\sqrt{2}}$. Find $b$.

Answer 1

$\cos \theta +\sin \theta =\sqrt{2\cos }(\theta -\pi /4)$
Substitute $(\theta -\pi /4)=x$
$\underset{x\to 0}{lim}\sqrt{2}\frac{(1-\cos x)}{16x^2}=\underset{x\to 0}{lim}\frac{2\sqrt{2}}{16}\frac{{\sin }^2\left(x/2\right)}{x^2}$
$=\frac{1}{4\sqrt{2}}.\frac{1}{4}.1=\frac{1}{16\sqrt{2}}$
Ans: 16