Can a limit of an integral be moved inside the integral?

Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.

The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.

The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+\frac{x}{n})^{-n}$, then $$\lim_{n\to +\infty}f_n(x)=e^{-x}=f(x)$$ We need to show that on $[0,1]$, $$\left\|f_n-f \right\|_{\infty}\to 0$$ But $$\left\|f_n-f \right\|_{\infty}=\sup_{x\in [0,1]}\left|f_n(x)-f(x)\right|= \sup_{x\in [0,1]}\left|(1+\frac{x}{n})^{-n}-e^{-x}\right| $$ We need to determine the maximum of $g_n(x)=(1+\frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.

$g_n(0)=0$, $g_n(1)=(1+\frac1n)^{-n}-e^{-1}$ and $$g_n^{\prime}(x_0)=0\Leftrightarrow (1+\frac{x_0}{n})^{-n-1}=e^{-x_0}$$ Whenever the latter is true, $g_n(x_0)=\frac{x_0}{n}e^{-x_0}$. Therefore, $$\left\|f_n-f \right\|_{\infty}=\max_{x\in [0,1]}\left|g_n(x)\right|=\max\left\{g_n(0),g_n(1),g_n(x_0)\right\}=\max\left\{0,(1+\frac1n)^{-n}-e^{-1},\frac{x_0}{n}e^{-x_0}\right\}\to 0 $$ as $n\to +\infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done