Prove that $f_n$ converges uniformly to $f$. For each $n \in \mathbb{N}$ define the function $f_n : [0, 1] \to \mathbb{R}$ by $f_n(x) = xe^{x/n}$, and define $f : [0, 1] \to \mathbb{R}$ by $f(x) = x$. Prove that $f_n \to f$ uniformly on $[0, 1]$.
So far I have:
$f_n(x) = xe^{x/n} \le e^{1/n}$. I was thinking of $N = \lceil 1/\epsilon \rceil$ and then:
$|f_n(x) - f(x)| = |xe^{x/n} - x| \le e^{1/n} \le e^{1/N} < \epsilon$, but I don't think $e^{1/N} < \epsilon$ is true. I have an exceedingly tough time choosing $N$ every time I have to prove one of these.
Not sure if I approached this correctly.

$|f(x)-f_n(x)| = |x||e^{x \over n}-1| \le e^{1 \over n}-1$.
