Question

$\underset{x\to 0}{lim}\frac{x\cot \left(4x\right)}{{\sin }^{2}x{\cot }^2\left(2x\right)}$ is equal to :

• A. $2$
• B. $0$
• C. $1$
• D. $4$

Answer 1

The correct option is C $1$

$\underset{x\to 0}{lim}\frac{x\cot \left(4x\right)}{{\sin }^{2}x{\cot }^2\left(2x\right)}$

$=\underset{x\to 0}{lim}\frac{x\cdot \frac{\cos 4x}{\sin 4x}}{{\sin }^{2}x\cdot \frac{{\cos }^{2}2x}{4{\sin }^{2}x{\cos }^{2}x}}$

$=\underset{x\to 0}{lim}\frac{4x\cdot \cos 4x\cdot {\cos }^{2}x}{\sin 4x\cdot {\cos }^{2}2x}$

$=\underset{x\to 0}{lim}\left(\frac{4x}{\sin 4x}\right)\cdot \left(\frac{\cos 4x{\cos }^{2}x}{{\cos }^{2}2x}\right)=1\times 1=1$