Does ${f_n}$ converge uniformly to $f$ on [-1,1]? Suppose $$f_n(x) = \frac{\sin(nx)+2nx^3}{(n+1)x^4+6}.$$ Does ${f_n}$ converge uniformly to $f$ on [-1,1] ?
How can I approach this question?
 A: First you have to find the limit function $f(x)$ on $[-1,1]$ by taking $n\to\infty$. Note that $f_n(0)=0$ for all $n$, so $f(0)=0$. If $x\ne 0$, then
$$\lim_{n\to\infty}\frac{\sin(nx)+2nx^3}{(n+1)x^4+6}=\lim_{n\to\infty}\frac{2nx^3\left(1+\frac{\sin(nx)}{2nx^3}\right)}{(n+1)x^4\left(1+\frac{6}{(n+1)x^4}\right)}=\lim_{n\to\infty}\frac{2nx^3}{(n+1)x^4}=\frac{2}{x}$$
so $f(x)=\frac 2x$ for $x\in[-1,0)\cup(0,1]$. As @Riley suggested in the comments, $f_n(x)$ are continuous (in fact, just bounded suffices) in $[-1,1]$ for all $n$. If we had uniform convergence, $f(x)$ would also be continuous (bounded) in that interval. However, $f(x)$ is discontinuous (unbounded) at $0$. 
A: The other answer is the quick way, but if you do not know the theorem on uniform limits of continuous functions, you can argue like this: 
The negation of uniform convergence is as follows: there is an $\epsilon>0$ such that for all integers $N$ there is an integer $n\ge N$ and an $x_n\in [-1,1]$ such that $|f_n(x_n)-f(x_n)|>\epsilon.$
Take $x_n=\frac{1}{n}$. Then, 
$\left|\frac{2}{x_n}-\frac{\sin(nx_n)+2nx_n^3}{(n+1)x_n^4+6}\right|=\left|2n-\frac{\sin 1+\frac{2}{n^2}}{\frac{n+1}{n^4}+6}\right|\ge 2n-\frac{\sin 1+2}{6}\ge 2n-\frac{1+2}{6}=2n-\frac{1}{2}$
So we may take $\epsilon=1.$
