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
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