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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.

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    What are you trying to show? A sequence cannot be "uniformly continuous". Perhaps you meant "uniformly convergent"?2012-12-07
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    You are correct and the correction has been noted.2012-12-07
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    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.2012-12-07
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    Sorry it took me a while to get back to this. My interval is [0,1].2012-12-08

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