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?
Best Answer
Alternative approach (without uniform convergence): let $t= \frac{x}{n}$ then $$\begin{align*}\int_0^1 \left(1 + \frac{x}{n}\right)^ndx&= n\int_0^{1/n} (1+t)^ndt\\ &=n\left[\frac{(1+t)^{n+1}}{n+1}\right]_0^{1/n} \\&= \frac{n}{n+1}\left(\left(1+\frac{1}{n}\right)^{n+1}-1\right)\\&\to e-1. \end{align*}$$