Question

Evaluate $\underset{x\to 0}{lim}{\left[\frac{(1+x{)}^{1/x}}{e}\right]}^{1/x}$

• A. $e^{-\frac{1}{2}}$
• B. $e^{-\frac{1}{2}}/2$
• C. $e^{\frac{1}{2}}/2$
• D. $e^{\frac{1}{2}}$

Answer 1

Answer: (A)

$e^{-\frac{1}{2}}$

Let $l=\underset{x\to 0}{lim}{\left[\frac{(1+x{)}^{1/x}}{e}\right]}^{1/x}$
Since $l>0$ taking log both sides,
$\log l=\underset{x\to 0}{lim}\frac{\frac{1}{x}\log (1+x)-1}{x}$
$\Rightarrow \log l=\underset{x\to 0}{lim}\frac{\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}-....)-1}{x}=-\frac{1}{2}$
$\therefore l=e^{-1/2}$
Hence correct choice is option 'A'