Can a sequence of unbounded functions be uniformly convergent? If I say have a sequence of functions $\displaystyle x^3 + 5\frac{x}{n}$.
I know that I can say it is point-wise convergent as $n$ goes to infinity to the limit function $x^3$ but can I simply say that as it isn't bounded it CAN NOT be uniformly convergent?
 A: Yes, you can have a sequence of unbounded functions that converges uniformly to an unbounded function.
Consider $f(x) = x^{2}$.  Given some $\epsilon > 0$, imagine drawing a bubble around $x^{2}$ of radius $\epsilon$.
Now let $f_{n}(x) = x^{2} + \frac{1}{n}$.  So $f_{n}$ is just $f$ shifted upward by $\frac{1}{n}$.
Clearly, $f_{n}$ converges to $f$ pointwise.  But look at the bubble around $x^{2}$ you drew.  Does there exist a point $N$ in the sequence $f_{n}$ such that for all points $n$ after $N$, the entire function $f_{n}$ is in the bubble around $f$?  The answer is yes, and hopefully you can see it if you drew a picture.  Because the answer is yes, $f_{n}$ converges uniformly to $f$, even though for all $n$, $f_{n}$ and $f$ are unbounded.
More generally, let $f(x)$ be any unbounded function and let $f_{n}(x)$ be defined as $f(x) - \frac{1}{n}$.  Then $f_{n}$ is also unbounded, and $f_{n}$ converges to $f$ pointwise (it is an excellent and easy exercise to prove this).
A: Any exercise is an opportunity to learn, provided you think about it a bit afterwards.  Excuse the pedagogical part of me that now emerges but I can't suppress it.


*

*One cannot say "uniformly convergent" and must always specify a set, so "uniformly convergent on $E$'' makes sense while the former doesn't.

*Our other friend who pointed out that the use of "it" was problematic is something you need to get used to as well.  Mathematicians are pretty fussy about language and referees will shred you to pieces over this in the future.

*The intent of the problem (since it is quite elementary) was to force you to apply the definition directly: $\epsilon$'s and $N$'s.  Even if you have a more elegant solution make sure you can also do this.

*Your instincts were absolutely right here.  You wanted to apply this principle: If $\{f_n\}$ converges uniformly on $E$ to a function $f$ and each $f_n$ has Property X on $E$ then $f$ must have property X on $E$.  This is valid if [Property X] is [bounded] or [unbounded] or [continuous] etc.  Not always though; it fails for [differentiable] for example.

*To make your solution work show that $x^3+ 5x/n$ converges uniformly to $x^3$ on some specified set $E$ if and only if $5x/n$ converges uniformly to zero on that set $E$.  If $E$ happens to be unbounded then this is impossible since the zero function is bounded while each function $5x/n$ is unbounded on such a set.
