Showing that a squence is not uniformly convergent Let $f_n(x)=\dfrac{\sin(nx)}{1+n^2x^2}$.  

First note that $f_n(x) \to 0$ pointwise.  Also, if you fix $\epsilon = \dfrac{1}{4}$ and let $x=\dfrac{1}{n}$, then $$|f_n(\frac{1}{n})-f(x)|=|
\frac{\sin(1)}{2}-0|=|\frac{\sin(1)}{2}|\geq \frac{1}{4}.$$

My question is, how much more should I show?  That is, I'm not sure how to make this argument more clearer.  I was marked off for lack of details.
The sequence converges pointwise because 
$\displaystyle \lim_{n\rightarrow +\infty}f_n(x_0)=\lim_{n\rightarrow +\infty}\frac{\sin(nx_0)}{1+n^2x_0^2}=0$.

To prove that the convergence is uniform, one must be show that $\displaystyle\lim_{n\rightarrow +\infty}\sup_{x\in[0,1]}\left|f_n(x)-f(x)\right|=0$, i.e. 

$$\sup_{x\in[0,1]}\left|\frac{\sin(nx)}{1+n^2x^2}-0\right| \to 0 \quad(n \to \infty).$$
The sequence converges pointwise because 
$\displaystyle \lim_{n\rightarrow +\infty}f_n(x_0)=\lim_{n\rightarrow +\infty}\frac{\sin(nx_0)}{1+n^2x_0^2}=0$.

To prove that the convergence is uniform, one must be show that $\displaystyle\lim_{n\rightarrow +\infty}\sup_{x\in[0,1]}\left|f_n(x)-f(x)\right|=0$, i.e. 

$$\sup_{x\in[0,1]}\left|\frac{\sin(nx)}{1+n^2x^2}-0\right| \to 0 \quad(n \to \infty).$$