Question

For $x>0,\underset{x\to 0}{\lim }\left[{\left(\sin x\right)}^{\frac{1}{x}}+{\left(\frac{1}{x}\right)}^{\sin x}\right]=$

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Find the value of $\underset{x\to 0}{\lim }\left[{\left(\sin x\right)}^{\frac{1}{x}}+{\left(\frac{1}{x}\right)}^{\sin x}\right]$

According to question

For $x>0$,

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

$=e^{-\infty }+e^{\underset{x\to 0}{\lim }(\frac{-{\displaystyle \frac{1}{x}}}{-\cos ecx*cotx})}$ $\left[\because x>0,\log \sin x=-\infty \right]$

$=0+e^{\underset{x\to 0}{\lim }(\frac{\tan x}{x}*\sin x)}$ $\left[\mathbf{\because }{\mathit{e}}^{\mathbf{-}\mathbf{\infty }}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\right]$

$=e^0$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{0}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\right]$

$=\mathbf{}\mathbf{1}$

Hence, Option ‘C’ is Correct.