Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{{\cot }^{3}x-\tan x}{\cos (x+\frac{\pi }{4})}$ is:

• A. $4\sqrt{2}$
• B. $4$
• C. $8\sqrt{2}$
• D. $8$

Answer 1

The correct option is D $8$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{{\cot }^{3}x-\tan x}{\cos (x+\frac{\pi }{4})}=\underset{x\to \frac{\pi }{4}}{lim}\frac{1-{\tan }^{4}x}{{\tan }^{3}x\left(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\right)}=\underset{x\to \frac{\pi }{4}}{lim}\frac{(1+{\tan }^{2}x)(1-{\tan }^{2}x)}{\frac{1}{\sqrt{2}}{\tan }^{3}x.(\cos x-\sin x)}=\sqrt{2}\underset{x\to \frac{\pi }{4}}{lim}\frac{(1+1).({\cos }^{2}x-{\sin }^{2}x).{\cos }^{3}x}{{\sin }^{3}x.{\cos }^{2}x.(\cos x-\sin x)}=2\sqrt{2}\underset{x\to \frac{\pi }{4}}{lim}\frac{(\cos x+\sin x)(\cos x-\sin x).\cos x}{{\sin }^{3}x.(\cos x-\sin x)}=2\sqrt{2}.\frac{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})\cdot \frac{1}{\sqrt{2}}}{{\left(\frac{1}{\sqrt{2}}\right)}^3}=2\sqrt{2}\cdot 2\sqrt{2}\cdot 1=8$