What is a good manner to see if a sequence of functions is uniform converging, when you're in the last stage of checking, i.e. finding an N that satisfies $|f_n(x) -f(x)| < \epsilon$?
Example: $f_n(x)=\frac{nx}{1+n^2x^2}$ on the domain $[-1,1]$
Last stage: $$|\frac{nx}{1+n^2x^2}|$$
I'd think that the next step is
$$|\frac{nx}{1+n^2x^2}|<|\frac{n}{1+n^2x^2}|$$
(Because 1 is the biggest value possible for $x$ and the second step being:
$$|\frac{nx}{1+n^2x^2}|<|\frac{n}{1+n^2x^2}| < |\frac{n}{1}|$$ (making the denominator as small as possible). I understand that $\frac{n}{1}$ will go to infinity and that is bigger than $\epsilon$, but I don't know in what kind of function sequences, it isn't possible to find an $x$ that satisfies all sequences. I only see that it doesn't converge uniform (and find it logical that it doesn't) after the teacher tells me.
I hope that my question is clear despite the broken English.
So you started out trying to prove an upper bound. If this worked out logically, then it would show uniform convergence. It didn't: you got an upper bound that doesn't go to zero.
So now the question is: does the sequence actually still converge uniformly, and your estimates were too sloppy, or should you use this as a hint that maybe the sequence doesn't actually converge uniformly? One great thing about using tight bounds, or at least nearly tight bounds, is that it automatically answers that question for us. In other words, if you had shown that $|f_n(x)| \leq n$ and in the process identified $x_n$ so that $|f_n(x_n)|=n$ (or $|f_n(x_n)| \geq n/2$ or something similar) then you would have just learned that the sequence doesn't converge uniformly, without having to start a whole separate calculation seeking out a lower bound. Sure, you'd still need to do that calculation to finish a rigorous proof, but you'd already know the answer. Figuring out the answer is often harder than proving that it is the answer.
The problem with your work, then, is that your bounds aren't tight at all. Your first bound is plenty tight: the numerator is $n$ at $x=\pm 1$. But the $x$'s where that bound is useful are all near the boundary. You then turn around and try to make the denominator small, but doing that requires you to go away from the boundary, which makes your previous estimate unhelpful.
So you need a different approach. Here's a hint: $f_n(x)=g(nx)$ where $g(y)=\frac{y}{1+y^2}$. So except at $x=0$, waiting for $f_n(x)$ to go to zero is equivalent to waiting for $nx$ to blow up. Does the size for $n$ required for that depend on $x$?