Question

If $f\left(x\right)=e^x$, then ${lim}_{x\to 0}{f\left(f\right(x\left)\right)}^{\frac{1}{\left(f\right(x\left)\right)}}$ is equal to (where {$x$} denotes fractional part of $x$)

• A. $f\left(1\right)$
• B. $f\left(0\right)$
• C. $0$
• D. Does not exist

Answer 1

The correct option is A $f\left(1\right)$

Given, $f\left(x\right)=e^x$

Then we need to find ${lim}_{x\to 0}f{\left[f\left(x\right)\right]}^{\frac{1}{\left(f\left(x\right)\right)}}=\_\_\_\_$

${lim}_{x\to 0}f{\left[f\left(x\right)\right]}^{\frac{1}{\left(f\left(x\right)\right)}}={lim}_{x\to 0}{\left[e^{e^x}\right]}^{\frac{1}{e^x}}$

$={lim}_{x\to 0}{e}^{(e^x\times \frac{1}{e^x})}$

$={lim}_{x\to 0}e$

$=e$ ....(1)

Given $f\left(x\right)=e^x$

Put $x=1$

$f\left(1\right)=e^1$

$=e$ ....(2)

From Equation (1) and (2), option A is correct.