Question: I would like to compute
$$\lim\_{n\to \infty }\int\_0^1 {\left(1+\frac{x}{n}\right)}^{-n}d x.$$

I know that $\lim\_{n\to \infty }(1+\frac{x}{n})^{-n}=e^{-x}$. I'm sure the convergence is uniform, but I have problem to show it. Let $$g\_n(x)=(1+\frac{x}{n})^{-n}-e^{-x}.$$

I have $$g\_n'(x)=-\left(1+\frac{x}{n}\right)^{-n-1}+e^{-x},$$
but how can I solve $g\_n'(x)=0$ and then find the maximum of $g'\_n$ ? I really don't see.

Answer: 
The sequence $(1+\frac{x}{n})^{-n}$ is a decreasingly monotone sequence of continuous functions which converges pointwise on the compact set $[0,1]$ to the function $e^{-x}$ which is also continuous. Therefore by [Dini'Theorem](https://en.wikipedia.org/wiki/Dini%27s_theorem) the convergence is also uniform.
There is e.g. the following elementary way: For $x\geq 0$, 
denote $g(x)=(1+x)^{-1} \in (0,1]$. Then for $n\geq 1$, $0\leq x\leq 1$ we get by the MVT:
$$ \left|g\left(\frac{x}{n}\right)-e^{-\frac{x}{n}}\right| \ \leq \ \left( \frac{1}{n}\sup\_{\xi\geq 0} \left| g'(\xi) +e^{-\xi} \right| \right)\ \ \left|\frac{x}{n}-0\right| 
 \ \leq \ \frac{1}{n} \times \frac{1}{n} $$
Using the expansion $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + b^{n-1})$ we obtain:
 $$\left| \left(1+\frac{x}{n}\right)^{-n}-e^{-x}\right|=\left|g\left(\frac{x}{n}\right)^n-(e^{-x/n})^n\right| \leq \frac{1}{n^2} 
\sum\_{k=1}^{n}g\left(\frac{x}{n}\right)^{k-1} (e^{-x/n})^{n-k}\leq \frac{1}{n^2} n =\frac{1}{n}$$