Question

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

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

Answer 1

The correct option is C

$1$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }{\left(\cos \left(x\right)\right)}^{{\cot }^2\left(x\right)}$

$\Rightarrow$$\cos {\left(0\right)}^{{\cot }^2\left(0\right)}$

$\Rightarrow 1^{\infty }$

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

Taking $\log$ both side,

$$\begin{gathered} \Rightarrow \ln \left(L\right)=\log \left(\underset{x\to 0}{\lim }{\left(\cos \left(x\right)\right)}^{{\cot }^2\left(x\right)}\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to 0}{\lim }\left(\log {\left(\cos \left(x\right)\right)}^{{\cot }^2\left(x\right)}\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to 0}{\lim }{\cot }^2\left(x\right)\log \left(\cos \left(x\right)\right) \\ \Rightarrow \ln \left(L\right)=\underset{x\to 0}{\lim }\frac{\log \left(\cos \left(x\right)\right)}{{\tan }^2\left(x\right)} \\ \Rightarrow \ln \left(L\right)=\frac{0}{0} \end{gathered}$$

Using L. Hospital rule,

$$\begin{gathered} \ln \left(L\right)=\frac{{\displaystyle \frac{1}{\cos \left(x\right)}}\left(-\sin \left(x\right)\right)}{{\sec }^2\left(2x\right).2} \\ \Rightarrow \ln \left(L\right)=\frac{0}{1.1.2} \\ \Rightarrow \ln \left(L\right)=0 \\ \Rightarrow L=e^0 \\ \Rightarrow L=1 \end{gathered}$$

Therefore the value of $\underset{x\to 0}{\lim }{\left(\cos \left(x\right)\right)}^{{\cot }^2\left(x\right)}$ is $1$.

Hence, option $\left(C\right)$ is the correct option.