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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?2012-03-10
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    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?2012-03-10
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    Think of rapidly oscillating sinusoidal functions that have small amplitudes.2012-03-10

2 Answers 2

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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 $x_0$.2012-03-10
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    @DidierPiau Yes. Thanks Didier.2012-03-10
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Put $$f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$$

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    How come you are able to find one?2012-03-10
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    @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.2012-03-10
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    @Victor: See Also: http://math.stackexchange.com/a/4672/1102. AD.: Done.2012-03-10
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    @Aryabhata Thanks for a nice link :)2012-03-10
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    @AD.: You are welcome :-)2012-03-10