Question

$\underset{x\to 0}{\lim }{(\cos ecx)}^{\frac{1}{\log \left(x\right)}}=$

• A. $0$
• B. $1$
• C. $\frac{1}{e}$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{e}$

Explanation for correct option:

Calculating the value of the limit:

Given expression is $\underset{x\to 0}{\lim }{(\cos ecx)}^{\frac{1}{\log \left(x\right)}}$

Simplifying the expression:

$$\begin{gathered} =\underset{x\to 0}{\lim }{(\csc x)}^{\frac{1}{\log \left(x\right)}} \\ =\underset{x\to 0}{\lim }{e}^{\frac{1}{\ln x}\ln (\csc x)}\mathbf{[}\mathbf{\because }{\mathbf{e}}^{\mathbf{l}\mathbf{n}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{a}\mathbf{x}\mathbf{]}\mathbf{} \\ =\underset{x\to 0}{\lim }{e}^{\frac{-\cot \left(x\right)}{{\displaystyle \frac{1}{x}}}}\mathbf{[}\mathbf{L}\mathbf{\text{'}}\mathbf{hospital}\mathbf{}\mathbf{rule}\mathbf{]} \\ =e^{\underset{x\to 0}{\lim }\left(\frac{{\displaystyle -\frac{1}{\tan x}}}{{\displaystyle \frac{1}{x}}}\right)} \\ =e^{-\underset{x\to 0}{\lim }\left(\frac{x}{\tan x}\right)}\left[\because \underset{x\to 0}{\lim }\frac{\mathrm{tanx}}{x}=1\right] \end{gathered}$$

Applying the limits:

$=e^{-1}$

Thus,

$\underset{x\to 0}{\lim }{(\cos ecx)}^{\frac{1}{\log \left(x\right)}}=e^{-1}=\frac{1}{e}$

Therefore, the correct answer is option (C).