Does the sequence $(f_n)$ on $[0, \infty)$ given by $ f_n(x) = > xe^{-nx} $ converge uniformly?
This is from Bartle's Elements of Real Analysis. I've already proven that the sequence is convergent (not uniformly) to the zero fucntion, $f$. But I'm guessing the answer to this question is "No" because $x$ is present in every approximation I come up with hence the choice of $m \in \Bbb N$ such that $\forall \epsilon \gt 0(n \ge m \implies |f_n(x) - f(x)| \lt \epsilon)$ depends on $x$. But I wish to prove this by presenting a subsequence $(f_k)$ and a corresponding subsequence $(x_k) \subseteq [0, \infty)$ such that $ |f_k(x) - f(x)| $ is not less than some positive number $\epsilon_{\circ}$. Any hints would be appreciated.
I am not allowed to use the series expansions of $e^x$. Just the definition $e^x = \lim \left({1 + \frac x n}\right)^n$ and maybe the logarithm-exponential relationships.
PS: The next exercise on the list is on the uniform convergence of $(x^2e^{-nx})$
$$(xe^{-nx})'=e^{-nx}-nxe^{-nx}=e^{-nx}(1-nx)$$
is $0$ only if $x=\frac{1}{n}$. So the maximum is attained at this point. So:
$$\lvert\lvert f_{n}\rvert\rvert_{C^{0}}=\frac{1}{ne}\to0$$
as $n$ tends to $\infty$.
Alternatively if you are aware that $e^{x}>x$ then for $x\ge0$ then:
$$xe^{-nx}=\frac{nx}{e^{nx}}\cdot\frac{1}{n}\le\frac{1}{n}$$
which holds for all $x\ge0$. This shows that the convergence is uniform.