TITLE: Computing $\lim_{x \to \infty} x \biggl[ \frac{1}{e} - \left( \frac{x}{x+1} \right)^x \biggr]$
QUESTION [6 upvotes]: I'm having some trouble solving finding the limit of:

$$\lim_{x \to \infty} x \left[ \frac{1}{e} - \left( \frac{x}{x+1} \right)^x \right]$$

I can see that $\left( \frac{x}{x+1} \right)^x = \left( 1 + \frac{-1}{x+1} \right)^x \rightarrow \frac{1}{e}$. That's why I thought the limit is 0. However, according to WolframAlpha it's $-\frac{1}{2e}$.
I tried to write $\frac{1}{e} - \left( \frac{x}{x+1} \right)^x$ as a series but I didn't find a way to... 
What's the right approach to find this limit?

REPLY [5 votes]: Let $t=\frac 1 x$ so by Taylor series we find easily
$$\lim_{x \to \infty} x \biggl[ \frac{1}{e} - \left( \frac{x}{x+1} \right)^x \biggr]=\lim_{t\to0}\frac1 t\left(\frac 1 e- \left(t+1\right)^{-\frac 1 t}\right)=\lim_{t\to0}\frac1 t\left(\frac 1 e- \exp\left(-\frac1 t\log(1+t)\right)\right)=\lim_{t\to0}\frac1 t\left(\frac 1 e- \exp\left(-\frac1 t(t-t^2/2+o(t^2))\right)\right)=\lim_{t\to0}\frac1 {et}\left(1- 1-\frac t 2\right)=-\frac{1}{2e}$$