How to compute this limit: $\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)$? I tried to compute a limit:
$$\lim_{x\to\infty}x\left(\frac
1e-\left(\frac x{x+1}\right)^x\right)$$
What I've done is that:
$$\begin{align}
&\lim_{x\to\infty}x\left(\frac
1e-\left(\frac x{x+1}\right)^x\right)\\
=&\lim_{x\to\infty}x\left(\frac
1e-\left(\frac x{x+1}\right)^{x+1}\left(\frac{x+1}{x}\right)\right)\\
=&\lim_{x\to\infty}\left(1-\frac 1{x+1}\right)^{x+1}x\left(1-\frac{x+1}{x}\right)\\
=&-\frac1e
\end{align}$$
But Wolfram|Alpha seems not to agree with me, in which it gave
$$-\frac1{2e}$$
So what actually does the error come from in my approach? And how should I compute the limit? Thanks in advance. 
 A: We'll make use of this standard limit :
$$\lim_{x\to0}\frac{e^x-1}{x}=1$$
Therefore, 
$$\lim_{x\to\infty}x\bigg(\frac{1}{e}-\bigg(\frac{x}{x+1}\bigg)^x\bigg)=\lim_{x\to\infty}\frac{\frac{1}{e}-\big(\frac{x}{x+1}\big)^x}{\frac{1}{x}}=\lim_{x\to\infty}\frac{e^{\ln\frac{1}{e}}-e^{x\ln\big(\frac{x}{x+1}\big)}}{\frac{1}{x}} = \lim_{x\to\infty}e^{x\ln\big(\frac{x}{x+1}\big)} \cdot\frac{e^{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}-1}{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}\cdot\frac{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}{\frac{1}{x}}$$
Which happens to be $$l = \frac{1}{e}\lim_{x\to\infty}\frac{\ln\frac{1}{e}-x\ln\big(\frac{x}{x+1}\big)}{\frac{1}{x}} = \lim_{y\to0_+}\frac{1}{e}\frac{y\ln\frac{1}{e}-\ln\frac{1}{1+y}}{y^2}=\lim_{y\to0_+}\frac{1}{e}\frac{\ln(1+y)-y}{y^2}=-\frac{1}{2e}$$
The last limit I guess you could solve using the technique here : Are all limits solvable without L'Hôpital Rule or Series Expansion, but it's easier to just apply l'Hospital once and some simple fraction manipulations.
