Question

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

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

Answer 1

Answer: (C)

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