Evaluate the limit with Taylor series How one can evaluate following limit: 
$$\lim_{x\to\infty} x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)?$$
I've found this exercise in the chapter about Taylor series, but I have no idea how to solve it.   
 A: Let $x=1/h$ i.e as $x\rightarrow \infty$, $1/h\rightarrow 0$. The limit is then
$$\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-\left({1+h}\right)^{-1/h}\right)=\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-e^{-\frac{1}{h}\ln\left({1+h}\right)}\right)$$
Also,
$$\ln(1+h)=h-\frac{h^2}{2}+\frac{h^3}{3}-\cdots \Rightarrow -\frac{1}{h}\ln(1+h)=-1+\frac{h}{2}-\frac{h^2}{3}+\cdots$$
$$\Rightarrow e^{-\frac{1}{h}\ln(1+h)}=e^{-1}e^{\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)}=\frac{1}{e}\left(1+\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)+\frac{\left(\frac{h}{2}-\frac{h^2}{3}+\cdots\right)^2}{2!}+\cdots\right)$$
Collecting only the terms linear in $h$, 
$$e^{-\frac{1}{h}\ln(1+h)}=\frac{1}{e}\left(1+\frac{h}{2}+\cdots \right)$$
Plugging this in the expression we have to evaluate,
$$\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-e^{-\frac{1}{h}\ln\left({1+h}\right)}\right)=\lim_{h\rightarrow 0} \frac{1}{h}\left(\frac{1}{e}-\frac{1}{e}\left(1+\frac{h}{2}+\cdots\right)\right)$$
$$=\lim_{h\rightarrow 0}\frac{-1}{2e}+(\text{terms containing h})=\boxed{\dfrac{-1}{2e}}$$
A: Using the Taylor series for $\log(1+x)$ and $e^x$,
$$
\begin{align}
\left(\frac{x}{x+1}\right)^{\large x}
&=\left(1+\frac1x\right)^{\large-x}\\[6pt]
&=e^{-x\log(1+\frac1x)}\\[6pt]
&=e^{-x\left(\large\frac1x-\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)}\\[9pt]
&=e^{-1}\,e^{\large\frac1{2x}+O\left(\frac1{x^2}\right)}\\[9pt]
&=\frac1e\left(1+\frac1{2x}+O\left(\frac1{x^2}\right)\right)\tag{1}
\end{align}
$$
Applying $(1)$, we get
$$
\begin{align}
\lim_{x\to\infty}x\left(\frac1e-\left(\frac{x}{x+1}\right)^{\large x}\right)
&=\lim_{x\to\infty}\left(-\frac1{2e}+O\left(\frac1x\right)\right)\\
&=-\frac1{2e}\tag{2}
\end{align}
$$
A: Write
$$
\frac1e-\left(\frac{x}{x+1}\right)^x=\left(e^{\frac1x}\right)^{-x}-\left(1+\frac{1}{x}\right)^{-x}
$$
By the mean value theorem,
$$
a^{-x}-b^{-x}=(-x)c^{-x-1}(a-b)
$$
with some $c$ between $a$ and $b$. Now
$$
a(x)-b(x)=e^{\frac1x}-1-\frac1x=\frac1{2x^2}+\frac1{6x^3}+...,
$$
$c(x)\to1$ and $c(x)^x\to e$ by the squeeze lemma and thus
$$
x\left(\frac1e-\left(\frac{x}{x+1}\right)^x\right)
=-\frac1{e+o(\frac1x)}\cdot \left(\frac12+\frac1{6x}+O\left(\frac1{x^2}\right)\right)\to-\frac1{2e}
$$
