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How to find a sequence of differentiable functions${f_n}$ with limit $0$ for which $\{f'_n\}$ diverges?

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You mean you need to find such a sequence? Well, what have you tried to do? –  William Mar 10 '12 at 18:37
    
Finding an answer to real analysis theorem sound like fun. Can you please get more clear with what you speak? You'd like to answer a question, but well to help in that process is the purpose of this site. So, your title reveals no information about the problem. Can you also tell us what have you tried? –  user21436 Mar 10 '12 at 18:56
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Think of rapidly oscillating sinusoidal functions that have small amplitudes. –  David Mitra Mar 10 '12 at 18:59

2 Answers 2

up vote 2 down vote accepted

How about $$ f_n (x) := \frac{\sin (n^2 x) }{ n }$$

Then the pointwise limit is $$ \lim_{n \to \infty} \left | f_n(x_0) \right | = \lim_{n \to \infty} \frac{\left | \sin (n^2 ) \right |}{\left | n \right |} \leq \lim_{n \to \infty} \frac{1}{|n |} = 0$$

But $$ f_n^\prime (x) = \frac{n^2 \cos (n^2 x)}{ n } = n \cos (n^2 x)$$

Whose pointwise limit diverges at the points $x_0 = 2 \pi k$ since

$$ \lim_{n \to \infty} n \cos (n^2 x_0) = \lim_{n \to \infty} n$$

Hope this helps. I thought I'd post this anyway even though I'm using the same function as AD in their answer. But I was typing this when their answer appeared and I see no reason to delete mine.

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The relation $|\cos(n^2x_0)|=|\cos(x_0)|$ for every $n\geqslant1$ holds only for very specific values of $x_0$. –  Did Mar 10 '12 at 20:05
    
@DidierPiau Yes. Thanks Didier. –  Matt N. Mar 10 '12 at 20:30

Put $$f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$$

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How come you are able to find one? –  Victor Mar 10 '12 at 19:18
    
@Victor The derivative measures tangents, an other ingredient that is a nice thing to know is that $x\mapsto f(nx)$ will look like $f$ but will be squeezed. –  AD. Mar 10 '12 at 21:30
    
@Victor: See Also: math.stackexchange.com/a/4672/1102. AD.: Done. –  Aryabhata Mar 10 '12 at 21:41
    
@Aryabhata Thanks for a nice link :) –  AD. Mar 10 '12 at 21:42
    
@AD.: You are welcome :-) –  Aryabhata Mar 10 '12 at 21:42

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