Does $f_n(x)=e^{-nx}$ converge uniformly to the zero function on the interval $(0,1]$? Consider $f_n(x)=e^{-nx}$.
I am wondering whether it converges uniformly to the zero function on the interval $(0,1].$ I can prove that it certainly converges uniformly to the zero function on any interval $[b,1]$ where $b>0.$ But I am just not sure about the case $(0,1].$
 A: You are right ! $(f_n)$ does not converge uniformly to the zero function on $(0,1]$. This can be seen by
$f_n(1/n)=1/e$  for all $n \in \mathbb N$
A: Hint: set $x_n\stackrel{\rm def}{=}\frac{1}{n}\in(0,1]$ for all $n\geq 1$. 
What is $f_n(x_n)$? What does that tell you about $\sup_{x\in(0,1]} \lvert f_n(x)\rvert$?
A: No. For if it did, then it would be the case that for all $\varepsilon > 0$ there exists $N\in \mathbb{N}$ such that $n>N$ implies that $f_n(x)=e^{-nx} < \varepsilon$ for all $x\in (0,1]$. However, let $\varepsilon = 0.5$, and given $n$, let $x_n = \frac{\ln (4/3)}{n}\in (0,1]$. Then $f_n(x_n)=0.75 > 0.5$. This is to say, there exists $\varepsilon$ such that for all $N$ we have an $m>N$ so that there exists $x$ such that $f_m(x) > \varepsilon$. Thus, the sequence does not converge uniformly. 
A: This is how I approach uniform convergence problems:
No matter how large $n$ be, there will be some $x$, close to zero, so that $-nx$ is still small in size, and thus $exp(-nx)$ is not close to zero, yet. This proves that for different $x$ the convergence happens at a different $n$ -- i.e. not uniformly converging.
