How to find a sequence of differentiable functions${f_n}$ with limit $0$ for which $\{f'_n\}$ diverges?
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1You mean you need to find such a sequence? Well, what have you tried to do? – Mar 10 '12 at 18:37
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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? – Mar 10 '12 at 18:56
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1Think of rapidly oscillating sinusoidal functions that have small amplitudes. – David Mitra Mar 10 '12 at 18:59
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Put $$f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$$
AD - Stop Putin -
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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. – AD - Stop Putin - Mar 10 '12 at 21:30
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@Victor: See Also: http://math.stackexchange.com/a/4672/1102. AD.: Done. – Aryabhata Mar 10 '12 at 21:41
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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.
Rudy the Reindeer
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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
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