Find the limit $\lim_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right)$ Find $$\lim_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right).$$
It first occurred to me that maybe taking the natural log, namely $\ln$, would do some help. Then the two terms immediately prevents me from doing so, because I can't separate these two terms, $\frac{x^{1+x}}{(1+x)^x}$ and $\frac{x}{e}$, under the operation of $\ln$.
I have no idea now. Anyone could help?
 A: Write that
$$
\frac{x^{1+x}}{\left(1+x\right)^{x}}=x\left(\frac{x}{1+x}\right)^{x}=x\left(1-\frac{1}{1+x}\right)^{x}=xe^{x \ln \left(1-\frac{1}{1+x}\right)}
$$
See if you can continue from here
A: Note that
$$\begin{align}\frac{x^{x}}{(1+x)^x}&=\left(1+\frac{1}{x}\right)^{-x}=\exp\left(-x\ln\left(1+\frac{1}{x}\right)\right)\\&=\exp\left(-x\left(\frac{1}{x}-\frac{1}{2x^2}+o(1/x^2)\right)\right)\\&=e^{-1}\exp\left(\frac{1}{2x}+o(1/x)\right)
\end{align}$$
where we used the Taylor expansion of $\ln(1+t)$ at $t=0$.
Hence, as $x\to +\infty$,
$$\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}=\frac{1}{e}\cdot\frac{\exp\left(\frac{1}{2x}+o(1/x)\right)-1}{1/x}.$$
Are you able to find the limit now?
P.S. We may also let $t=1/x$ and apply L'Hopital (twice) to
$$\begin{align}
\lim_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right)&=\lim_{t\to 0^+}\frac{\exp\left(-\ln\left(1+t\right)/t\right)-1/e}{t}\\
&\stackrel{H}{=}
\lim_{t\to 0^+}\left(
\left(\frac{\ln(1+t)}{t^2}-\frac{1}{t(1+t)}\right)\exp(-\ln(1+t)/t)
\right)
\\
&=
\frac{1}{e}\cdot\lim_{t\to 0^+}
\frac{\ln(1+t)-\frac{t}{1+t}}{t^2}\\
&\stackrel{H}{=}
\frac{1}{e}\cdot\lim_{t\to 0^+}
\frac{\frac{1}{1+t}-\frac{1}{(1+t)^2}}{2t}.\end{align}$$
A: One may write, as $x \to \infty$,
$$
x\ln\left(1+\frac1x \right)=x\left(\frac1x-\frac1{2x^2} +o\left(\frac1{x^2} \right)\right)
$$ and
$$
\frac{x^x}{(1+x)^x}=\left(1+\frac1x\right)^{-x}=e^{\large-x\ln\left(1+\frac{1}{x}\right)}
$$ to conclude.
A: I would write
$$\frac{x^{1+x}}{(1+x)^x}= x \frac{x^{x}}{(1+x)^x} = x \left(\frac{1}{1+1/x}\right)^x.$$
Then
$$\ln \left(\frac{1}{1+1/x}\right)^x = -x \ln(1+1/x)=-1+1/2x + o(1/x^2)$$ and
$$\frac{x^{1+x}}{(1+x)^x} = \frac{x}{e}\left(1+1/(2x) + o(1/x^2) \right)=\frac{x}{e}+1/(2e) + o(1/x).$$
Therefore
$$\lim_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right) = 1/(2e)$$
