Let $f_n(x)=xne^{-nx}$ for all $x \ge 0$ and $n \ge 1$. Show that $(f_n)$ converges to zero on $[0, \infty)$ pointwise but not uniformly Let $f_n(x)=xne^{-nx}$ for all $x \ge 0$ and $n \ge 1$. Show that $(f_n)$ converges to zero on $[0, \infty)$ pointwise but not uniformly. 
I know we have to evaluate at $f_n(0)$ and where $x$ does not equal $0$ then take the limit of it but I don't know for sure if we can separate the limits so it would be $\lim x \cdot \lim n \cdot \lim e^{-nx}$.
 A: Hint: It suffices to show that $\lim_{n \to \infty} n e^{-nx} = 0$ for any positive $x$. For this you can use L'Hopital's rule, differentiating with respect to $n$. To show the convergence is not uniform, take $x = 1/n$ and use this to show convergence is not uniform.
A: Pointwise convergence is straightforward since for any $x\in\mathbb{R}^+$
$$\lim_{n\to +\infty}f_n(x) = \lim_{z\to +\infty} ze^{-z}= 0, $$
but since $f_n(x)\geq 0$ and:
$$\int_{0}^{1/n}f_n(x)\,dx = \frac{e-2}{e}\cdot\frac{1}{n}$$
for any $n\in\mathbb{N}^*$ there exists $y\in[0,1/n]$ such that $f_n(y)>\frac{e-2}{e}>0$, contradicting uniform convergence. We even get a stronger bound by noticing that:
$$ f_n\left(\frac{1}{n}\right) = \frac{1}{e} > 0.$$
A: Pointwise convergence is due to
$$
|f_n(x)-0|=|xne^{-nx}|={nx\over e^{nx}}\to 0\text{ as }n\to \infty\text{ for each fixed }x\in [0,\infty).$$
On the other hand, $f_n\to 0$ uniformly on $[0,\infty)$ requires $\sup_{x\in[0,\infty)}|f_n(x)-0|\to 0$ as $n\to \infty$, but here $$\sup_{x\in[0,\infty}|f_n(x)-0|=\sup_{x\in[0,\infty)}|xne^{-nx}|={1\over e}\not\to 0\text{ as }n\to\infty.$$
