Let $f_n(x)= \frac x{x+n}$ for $n \in \Bbb{N}$. Show that if $a>0$ then $f_n$ converges to 0 uniformly on $[0,a]$ and show that the convergence is not uniform on $[0,\infty]$.
So I've deduced that I need to somehow reduce the fraction into $x/n$ which goes to $0$ as $n$ goes to infinity. But I don't know how to show the convergence isn't uniform until infinity. Couldn't I just make $a$ really large? Why would it converge uniformly for a large $a$ but not an infinite $a$?
Let $a>0$ and choose $\varepsilon>0$ (we may assume that $\varepsilon<1$). We want to show that $\frac{x}{x+n}<\varepsilon$ for all $x$ and for all $n$ greater than some $N$. Take $N$ equal to the first positive integer greater than $\frac{a-\varepsilon a}{\varepsilon}$. Then for all $n\geq N$ we have that $n\geq \frac{1}{\varepsilon}(1-\varepsilon)a\geq \frac{1}{\varepsilon}(1-\varepsilon )x$ for all $x\in [0,a]$. Or equivalently, $\frac{x}{x-n}<\varepsilon$. Clearly this argument uses that $a<\infty$. Also see Umberto P.'s answer to prove yourself that it fails when $a=\infty$.