Question

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

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

Answer 1

Answer: (C)

$\underset{x\to \infty }{\lim }{\left[\frac{e}{(1+1/x{)}^x}\right]}^x=\underset{y\to 0}{\lim }{\left[\frac{e}{(1+y{)}^{1/y}}\right]}^{1/y}$
$=\underset{y\to 0}{\lim }{e}^{\frac{1}{y}\ln \left[\frac{e}{(1+y{)}^{y_y}}\right]}$
Now, we have, $\underset{y\to 0}{\lim }\frac{1}{y}\ln \left[\frac{e}{(1+y{)}^{1/y}}\right]$
$=\underset{y\to 0}{\lim }\frac{\ln e-\frac{1}{y}\ln (1+y)}{y}$
$=\underset{y\to 0}{\lim }\frac{y-\ln (1+y)}{y^2}$
$=\underset{y\to 0}{\lim }\frac{y-\left(y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\dots \right)}{y^2}$
$=\underset{y\to 0}{\lim }\frac{1}{2}-\frac{y}{3}+\frac{y^2}{4}-\dots =\frac{1}{2}$
Hence, the required limit is $e^{1/2}$.