For a sequence $f_{k}(x)=(\frac{x}{k})e^{-\frac{x}{k}},\ x\in[0,\infty)$, does this sequence uniform convergent For a sequence $f_{k}(x)=(\frac{x}{k})e^{-\frac{x}{k}},\ x\in[0,\infty)$, does this sequence uniform convergent in the given interval? If not, does it uniform convergent in a slightly smaller interval? 
I have no idea how to deal with the limit of the sequence? Does any one can help me? Thanks!
 A: From what I can tell, it appears that the sequence approaches $f(x)=0$, since, for every fixed $x\in[0,\infty)$, $\left(\frac{x}{k}\right)\rightarrow0$ and $e^{-x/k}\le1$.  Thus, to show that it's uniformly convergent, we want to show that, given $\epsilon>0$, there exists an $K\in\mathbb{N}$ such that, whenever $k\ge K$, $|f_k(x)-0|<\epsilon$, for every $x\in[0,\infty)$. Somewhat informally speaking now, we know that $e^{-x/k}\le1$ and so if we can find when $\left(\frac{x}{k}\right)<\epsilon$, then we should be good.  This would motivate us to pick a $K>\frac{x}{\epsilon}$.  However, that means that our choice of $K$ depends on $x$, and so the sequence of function only converges pointwise on $[0,\infty)$, not uniformly.
Now if you were to consider the interval $[0,b]$ for any fixed $b>0$, then that gives you an upper bound on $x$, and so you could choose $$K>\frac{b}{\epsilon}\ge\frac{x}{\epsilon},$$
which gives you a $K$ not dependent on $x$, and so it would uniformly converge on $[0,b].$
A: Let us look at the maximum of $f_k$ over its domain. One has
$$f'_k(x)=e^{-\frac{x}{k}}\left(\frac{1}{k}-\frac{x}{k^2}\right)$$
And this is null for $x=k$ and therefore
$$\sup_{x\in [0,\infty)}f_k(x)=e^{-1}$$
And therefore the sequence is not uniformly convergent. By the Heine theorem it is uniformly convergent in any closed and bounded interval
A: This sequence converges uniformly on every interval of the form $[0,a]$ to $0$, because we have
$$0\le f_k(x)\le \frac{a}{k}\longrightarrow_{k\to\infty}0$$
The convsrgence is not uniform on $\mathbb{R}^+$, since $f_k(k)=e^{-1}$ for every $ k$.
