Question: The figure shows the cross section of 
 a swimming pool. What angle does the slopping bottom make with the horizontal?

[![enter image description here](https://i.stack.imgur.com/h8A8Y.png)](https://i.stack.imgur.com/h8A8Y.png)
how can i solve that

Answer: 
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
$$\begin{align}
\lim\_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right)&=\lim\_{t\to 0^+}\frac{\exp\left(-\ln\left(1+t\right)/t\right)-1/e}{t}\\
&\stackrel{H}{=}
\lim\_{t\to 0^+}\left(
\left(\frac{\ln(1+t)}{t^2}-\frac{1}{t(1+t)}\right)\exp(-\ln(1+t)/t)
\right)
\\
&=
\frac{1}{e}\cdot\lim\_{t\to 0^+}
\frac{\ln(1+t)-\frac{t}{1+t}}{t^2}\\
&\stackrel{H}{=}
\frac{1}{e}\cdot\lim\_{t\to 0^+}
\frac{\frac{1}{1+t}-\frac{1}{(1+t)^2}}{2t}.\end{align}$$
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
$$\lim\_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right) = 1/(2e)$$