Uniform convergence of $f_n(x) = \left(1 + \frac{x}{n}\right)^n$ when calculating limit Calculate$$
\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx
$$
My attempt - if
$$
f_n(x) = \left(1 + \frac{x}{n}\right)^n
$$
converged uniformly for all $x \in [0,1]$ then I could swap integral with limes and solve it:
$$
\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx = 
 \int_0^1 \lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^ndx =
 \int_0^1 e^x dx = e^x|_{0}^{1} = e - 1
$$
I must then prove that $f_n(x)$ is indeed uniformly convergent. I already know that
$f_n(x) \rightarrow e^x$. If $f_n(x)$ converges uniformly then for each epsilon the following statement must hold
$$
\sup_{x \in [0,1]} \left|f_n(x) - f(x)\right| < \epsilon
$$
How can I prove this?

You can use Dini's theorem.
On a compact set $K$, if a sequence of continuous functions $\langle f_n(x) \rangle$
$a)$ is monotone in $n$ for each $x \in K$
$b)$ converges pointwise to a continuous function of $x \in K$
then the convergence is uniform.
