Question

The value of $\underset{x\to 0}{lim}(\cos x{)}^{{\cot }^{2}x}$ is

• A. $e^{-1}$
• B. $e^{-1/2}$
• C. $1$
• D. Does not exist

Answer 1

The correct option is B $e^{-1/2}$

Let $L=\underset{x\to 0}{lim}(\cos x{)}^{{\cot }^{2}x}$

Taking $\log$ on both sides

$\therefore \log L=\log \underset{x\to 0}{lim}(\cos x{)}^{{\cot }^{2}x}$

$\log lim-\underset{x\to 0}{lim}\log (\cos x{)}^{co{t}^{2}x}$

$\log L=\underset{x\to 0}{lim}{\cot }^{2}x\log \left(\cos x\right)$

$\log L=\underset{x\to 0}{lim}\frac{{\cot }^{2}x\log \left(\cos x\right)}{{\sin }^{2}x}$

Applying L' hospital's rules

$\log L=\underset{x\to 0}{lim}\left[\frac{-{\cos }^{2}x.\sin x+\log \cos x(-2\cos x.\sin x)\cos x}{2\sin x.\cos x}\right]$

$\log L=\underset{x\to 0}{lim}\left[\frac{-\frac{1}{2}\sin 2x-\sin 2x\log \cos x\cos x}{\sin 2x}\right]$

$\log L=\underset{x\to 0}{lim}(-\frac{1}{2}-\log \cos x)$

$\log L=-\frac{1}{2}$

$\therefore L=e^{-1/2}$