Find $ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$ 
Find the limit of
$$ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$$

Let $$u= 1 +\frac{x}{n} \implies du =\frac{1}{n} dx \implies n \cdot du = dx$$
at $x=0$ $u=1$ and at $x=1$ $u=1+\frac{1}{n}$  so now limit will change from $1$ to $1+\frac{1}{n}$
Back to the integral
$$ \lim\limits_{n \rightarrow \infty} \left( n \cdot \int_{1}^{1+\frac{1}{n}} u^n du \right)= \lim\limits_{n \rightarrow \infty} \left( n \cdot \left[ \frac{nu^{n+1}}{n+1} \right]_1^{1+\frac{1}{n}} \right) = \lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1}  \left[  u^{n+1} \right]_1^{1+\frac{1}{n}} \right) $$
$$\implies\lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1}  \left[  \left(1+\frac{1}{n} \right)^{n+1}-1 \right] \right)=\infty$$
Is my finding correct? Is the procedure of taking the limit before completing the integration correct?
Much appreciated

Since
$\left(1+ \frac{x}{n}\right)^n
\to e^x
$,
$
\lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx
= \int_{0}^{1} e^x dx
=e-1
$.
