Question

If $\underset{x\to \infty }{lim}x\ln \left(e{(1+\frac{1}{x})}^{1-x}\right)=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, then the value of $m+n$ is

Answer 1

$L=\underset{x\to \infty }{lim}x\ln \left(\frac{e(1+\frac{1}{x})}{{(1+\frac{1}{x})}^x}\right)$
$\Rightarrow L=\underset{x\to \infty }{lim}x[1+\ln (1+\frac{1}{x})-x\ln (1+\frac{1}{x})]$
Put $\frac{1}{x}=t$
Then $L=\underset{t\to 0}{lim}\frac{1}{t}(1+\ln (1+t)-\frac{\ln (1+t)}{t})$
$=\underset{t\to 0}{lim}(\frac{\ln (1+t)}{t}+\frac{t-\ln (1+t)}{t^2})=1+\underset{t\to 0}{lim}\left(\frac{t-\ln (1+t)}{t^2}\right)\left(\frac{0}{0}\text{form}\right)$
Apply L'Hospital Rule
$L=1+\underset{t\to 0}{lim}\frac{1-\frac{1}{1+t}}{2t}$
$\Rightarrow L=1+\frac{1}{2}=\frac{3}{2}=\frac{m}{n}\Rightarrow m+n=3+2=5$