# 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.

-
What are you trying to show? A sequence cannot be "uniformly continuous". Perhaps you meant "uniformly convergent"? – Giuseppe Negro Dec 7 '12 at 0:49
You are correct and the correction has been noted. – emka Dec 7 '12 at 0:50
What interval are you working on? $I=[0,\infty)$? Presumeably, you were expected to finish by saying "and so there is no $N$ so that $|f_n(x)-0|<\epsilon$ for all $x\in I$ and $n\ge N$". Or perhaps you were expected to explain why the pointwise limit is the zero function. – David Mitra Dec 7 '12 at 1:34
Sorry it took me a while to get back to this. My interval is [0,1]. – emka Dec 8 '12 at 6:51