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: 
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
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
$$\lim\_{x\rightarrow +\infty}\left(\frac{x^{1+x}}{(1+x)^x}-\frac{x}{e}\right) = 1/(2e)$$