On which interval is $f_n = nx e^{-nx}$ is the pointwise limit of $f_n$ uniform convergent?

Question

Given is that $f_n(x) = nx e^{-nx}$ on $[0,\infty)$
I computed that the pointwise limit is $f(x)= 0$ for all $x \in[0,\infty)$
I would like to compute the interval in $[0,\infty)$ such on which $f_n(x)$ is uniform convergent.
So I began as follows:
$$\text{sup}_{x \in [0,\infty)} |f_n(x) - 0| \\ = f_n(\frac{1}{n}) \\ =\frac{1}{e}$$ So $f_n(x)$ is not uniformly convergent on the whole of $[0,\infty)$ since $$\text{sup}_{x \in [0,\infty)} |f_n(x) - 0| =\frac{1}{e} \nrightarrow 0$$
On which intervals is it then uniformly convergent and on which is it not and why?

Answer 1

Hint: $f_n$ is decreasing on $[\epsilon,\infty)$ if $\frac 1n < \epsilon$. Hence $\sup_{x\in [\epsilon,\infty)}f_n(x) = f_n(\epsilon)$.