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?

To show uniform convergence, you can just estimate the error term:
Note that
$$
n\log(1+x/n)=x-\frac1{2n}x^2+\frac1{3n^2}x^3-\dots
$$
is an alternating series with each term smaller in magnitude (since $x\in[0,1]$), so
$$
\lvert x-n\log(1+x/n)\rvert\leq\frac1{2n}x^2
$$
So for all $x\in[0,1]$, we have
\begin{align*}
0&\leq e^x-\left(1+\frac{x}{n}\right)^n\\
&=e^x-e^{n\log(1+x/n)}\\
&\leq\lvert x-n\log(1+x/n)\rvert\cdot\sup\{e^\xi\mid n\log(1+x/n)\leq\xi\leq x\} &&\text{(MVT)}\\
&\leq \frac{x^2}{2n}e^x\leq\frac{e}{2n}
\end{align*}
So the convergence is uniform.
