Question

$\underset{x\to \mathrm{\pi }/4}{\lim }\frac{4\sqrt{2}-{\left(\cos x+\sin x\right)}^5}{1-\sin 2x}$ is equal to

(a) $5\sqrt{2}$

(b) $3\sqrt{2}$

(c) $\sqrt{2}$

(d) none of these

Answer 1

(a)

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{4\sqrt{2}-{\left(\cos x+\sin x\right)}^5}{1-\sin 2x} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{2^{{\displaystyle \frac{5}{2}}}-{\left({\left(\cos x+\sin x\right)}^2\right)}^{{\displaystyle \frac{5}{2}}}}{2-\left(1+\sin 2x\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{2^{{\displaystyle \frac{5}{2}}}-{\left({\left(\cos x+\sin x\right)}^2\right)}^{{\displaystyle \frac{5}{2}}}}{2-{\left(\cos x+\sin x\right)}^2} \\ \mathrm{Let}t={\left(\cos x+\sin x\right)}^2 \\ x\to \frac{\mathrm{\pi }}{4} \\ \therefore t={\left(\cos \frac{\mathrm{\pi }}{4}+\sin \frac{\mathrm{\pi }}{4}\right)}^2\to {\left(\sqrt{2}\right)}^2=2 \\ =\underset{t\to 2}{\lim }\frac{2^{{\displaystyle \frac{5}{2}}}-{\left(t\right)}^{{\displaystyle \frac{5}{2}}}}{2-\left(t\right)} \\ =\frac{5}{2}{\left(2\right)}^{\frac{3}{2}}\left[\because \underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}=n{a}^{n-1}\right] \\ =5\sqrt{2} \end{gathered}$$