Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{16\sqrt{2}-(\sin x+\cos x{)}^9}{1-\sin 2x}$ equals

• A. $18\sqrt{2}$
• B. $12\sqrt{2}$
• C. $32\sqrt{2}$
• D. $36\sqrt{2}$

Answer 1

The correct option is D

$36\sqrt{2}$

$L=\underset{x\to \frac{\pi }{4}}{lim}\frac{16\sqrt{2}-(\sin x+\cos x{)}^9}{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}$ [because ${\sin }^{2}x+{\cos }^{2}x=1$ and $\sin 2x=2\sin x\cos x$]

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{16\sqrt{2}-(\sin x+\cos x{)}^9}{(\sin x-\cos x{)}^2}$

Put $x=\frac{\pi }{4}+t$

Then $\sin x=\sin (\frac{\pi }{4}+t)$

$=\sin \frac{\pi }{4}\cos t+\cos \frac{\pi }{4}\sin t$

$=\frac{1}{\sqrt{2}}\cos t+\frac{1}{\sqrt{2}}\sin t$

$=\frac{\sin t+\cos t}{\sqrt{2}}$

and $\cos x=\cos (\frac{\pi }{4}+t)$

$=\cos \frac{\pi }{4}\cos t-\sin \frac{\pi }{4}\sin t$

$=\frac{1}{\sqrt{2}}\cos t-\frac{1}{\sqrt{2}}\sin t$

$=\frac{\cos t-\sin t}{\sqrt{2}}$

$\Rightarrow \sin x-\cos x=\frac{\sin t+\cos t}{\sqrt{2}}-\frac{\cos t-\sin t}{\sqrt{2}}=\frac{2\sin t}{\sqrt{2}}=\sqrt{2}\sin t$

and $\sin x+\cos x=\frac{\sin t+\cos t}{\sqrt{2}}+\frac{\cos t-\sin t}{\sqrt{2}}=\frac{2\cos t}{\sqrt{2}}=\sqrt{2}\cos t$

$\Rightarrow L=\underset{t\to 0}{lim}\frac{16\sqrt{2}-(\sqrt{2}\cos t{)}^9}{(\sqrt{2}\sin t{)}^2}$

$=\underset{t\to 0}{lim}\frac{16\sqrt{2}-16\sqrt{2}{\cos }^{9}t}{(\sqrt{2}\sin t{)}^2}$

$=\frac{16\sqrt{2}}{2}\underset{t\to 0}{lim}\frac{1-{\cos }^{9}t}{{\sin }^{2}t}$

$=8\sqrt{2}\underset{t\to 0}{lim}\frac{1-{\cos }^{9}t}{{\sin }^{2}t}$

$=8\sqrt{2}\underset{t\to 0}{lim}\frac{1-\cos t}{t^2}\times \frac{t^2}{{\sin }^{2}t}\times (1+\cos t+{\cos }^{2}t+...+{\cos }^{8}t)$

$=8\sqrt{2}\times \frac{1}{2}\times 1\times 9=36\sqrt{2}$