Question

The value of $\underset{x\to \pi /4}{lim}\frac{8-\sqrt{2}(\cos x+\sin x{)}^5}{1-\sin 2x}$ is

Answer 1

Reqd. limit $=\sqrt{2}\underset{x\to \pi /4}{lim}\frac{4\sqrt{2}-(\cos x+\sin x{)}^5}{1-\sin 2x}$
$=\sqrt{2}\underset{x\to \pi /4}{lim}\frac{2^{5/2}-\left(\right(\sin x+\cos x{)}^2{)}^{5/2}}{1-\sin 2x}$
$=\sqrt{2}\underset{x\to \pi /4}{lim}\frac{2^{5/2}-(1+\sin 2x{)}^{5/2}}{2-(1+\sin 2x)}$
$=\sqrt{2}\underset{u\to 2}{lim}\frac{2^{5/2}-u^{5/2}}{2-u}$ ; $(u=1+\sin 2x)$
using L hospital's rule,
$=\sqrt{2}\frac{5}{2}{.2}^{3/2}=10$