Can you give me an example of a sequence which converges uniformly on $[0,L]$ $L>0$ but not on $[0,\infty]$ Can you give me an example of a sequence which converges uniformly on $[0,L]$ $L>0$ but not on $[0,\infty]$
I thought about $g_n(x)= \frac{nx^2}{n^3+x^2}$ Since $x \to \infty => g_n \to n$
And $\lim_{n \to \infty}g_n = 0 $
but when $x \to \infty$ And $\lim_{n \to \infty}g_n = \infty $ 
Which Makes the function $G(x)= \lim_{n \to \infty} g_n$ discontinuous. 
Therefore $g_n$ can't be uniformly bounded on $[0, \infty]$
What do you think? I believe I'm making a wrong approach but just wanted to share so I can spot some mistakes, thanks!
 A: (1). The  3rd line of your answer is gibberish but I think you have the right idea. For each real $x\geq 0$ we have $\lim_{n\to \infty}g_n(x)=0,$ but for each $n$ we have $\lim_{x\to \infty}g_n(x)=1.$ So for each $n$ there exists (some sufficiently large) real $x_n$ with $g_n(x_n)>1/2,$ so  $g_n$  does not converge uniformly to $0$ on $[0,\infty)$.
But for real $r>0$ and $x\in [0,r]$ we have $g_n(x)\leq nr^2/(n^3+x^2)\leq nr^2/n^3=(r/n)^2,$ so $g_n\to 0$ uniformly on $[0,r].$
(2). Caution: A sequence $(f_n)_n$ of bounded continuous functions may converge uniformly to $f$ on $[0,\infty)$ without any "continuity of $f$ at $\infty$". For example, for $m\in \mathbb N_0,$ let $f$ be linear on $[2m,2m+1]$ with $f(2m)=0$ and $f(2m+1)=1;$ and let $f$ be linear on $[2m+1,2m+2]$ with $f(2m+2)=0$. Now let $f_n(x)=f(x)+1/n.$ 
Continuity of a uniform limit of continuous functions is a theorem about  domains that are subsets of $\mathbb R$.
(3). Let $\|g_n-g\|_D=\sup_{x\in D}|g_n(x)-g(x)|.$ Then "$(g_n)_n$ converges uniformly to $g$ on $D$ as $n\to \infty\;$" can be said as: $\; \lim_{n\to \infty}\|g_n-g\|=0.$
In your example, with $D=[0,\infty)$ we have $g=0$ and $\|g_n-g\|_D=\|g_n\|_D\geq \lim_{x\to \infty}|g_n(x)|=1$ so convergence on $[0,\infty)$ is not uniform. 
A: 1)In your example,
$g_n$ converges point wise to $0$ on every $[0,L]$ where $L \gt 0$. But the sequence $g_n(n) = \frac {n^3}{n^3 + n^2} \to 1.$ Hence $g_n$ is not uniformly convergent on $[0,\infty)$. Also $\sup \{|g_n(x)-0| : 0 \le x \lt L\}= \sup \{ |\frac {nx^2}{n^3 + x^2}| : 0 \le x \lt L\} \le \sup\{ |(\frac xn)^2| : 0 \le x \lt L\}= \frac {L^2}{n^2} \to 0$ Hence $g_n$ is uniformly convergent on $[0,L]$ for $L \gt 0$.
2)As another example, take $g_n(x)=\frac {x}{x+n}$ on $[0,\infty)$.
Then $g_n(x) \to 0$ pointwise for all $x \in [0,
\infty)$.
But since $g_n(n) \to \frac 12$, $g_n$ is not uniformly continuous on $[0,\infty)$.
Now take any $L \gt 0$,
Then $\sup \{|g_n(x)-0| : 0 \le x \lt L\} \le \frac {L}{n} \to 0$.
Hence $g_n$ is uniformly convergent on $[0,L]$ for any $L \gt 0$.
A: A good and general example is given by the sequence of partial sums of an entire series of infinite convergence radius; for instance, with the exponential function: $f_{n}(x)=\sum_{k=0}^{n}{\frac{x^{k}}{k!}}$.
