Evaluation of limit

Question

$$\lim_{ x\to 0}x(\frac{1}{e}-(\frac{x}{1+x})^x)$$

How to proceed after applying LH rule once.

Answer 1

L'Hospital's rule is not the alpha and omega of limits computation. You just have to use Taylor's formula at order $2$.

First a change of variable: set $u=\dfrac1x$; $\;u\to 0\;$ as $x\to\infty$. Now \begin{align}\ln\Bigl(\frac{x}{1+x}\Bigr)^x&=x\ln\frac{x}{1+x}=\frac1u\ln\frac{1}{1+u}=\frac{-\ln(1+u)}{u}\\ &=\frac{-u+\dfrac{u^2}2+o(u^2)}u=-1+\frac u2+o(u). \end{align} Thus we have $$\Bigl(\frac{x}{1+x}\Bigr)^x=\mathrm e^{x\ln\!\tfrac{x}{1+x}}=\mathrm e^{-1+\tfrac u2+o(u)}=\frac{e^{\tfrac u2+o(u)}}{\mathrm e}=\frac{1+\dfrac u2+o(u)}{\mathrm e}=\frac 1e+\frac{u}{2\mathrm e}+o(u),$$ so $$x\biggl(\frac 1e-\Bigl(\frac{x}{1+x}\Bigr)^x\biggr)=\frac{-\dfrac{u}{2\mathrm e}+o(u)}{u}=-\dfrac1{2\mathrm e}+o(1)\to-\dfrac1{2\mathrm e}. $$