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Is the fucntion

$$ f(x) = \begin{cases} x\sin(\frac1x) & \text{for }x\ne0, \\ 0 & \text{for }x=0. \end{cases} $$ differentiable at $x=0$?

S L
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potato
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2 Answers2

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Hint

The function $x\mapsto \sin\left(\frac1x\right)$ doesn't have a limit at $0$ by considering the sequence $\left(\frac{1}{\frac\pi2+n\pi}\right)$

  • lol ... your hint is too complicated as usual to apply. – S L May 25 '14 at 09:07
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    For this reason I called it hint and not answer! The OP should do some effort;-) –  May 25 '14 at 09:10
  • "as usual"!! Oh I'll check my previous hints and apparently I have a bad reputation on it lol –  May 25 '14 at 09:18
  • you consfused DonAntonio here and now you confused me where is this hint applicable. It's a classic example I learnt when I was in High School. Probably it has duplicate somewhere. – S L May 25 '14 at 09:20
  • I think there's nothing wrong with this hint, but it complicates things a little bit. All you need to show divergence at $x_0 =0$ is to see that $\sin t$ oscillates between -1 and 1 as $t \to \infty$ – Alex May 25 '14 at 09:46
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    @Alex "oscillates between -1 and 1" isn't a rigorous argument and I think that to do mathematics we need sometimes a little bit of complication. –  May 25 '14 at 10:02
  • If the OP can't use the intuitive divergence of $\sin$ function, I guess this whole proof should be presented: http://math.stackexchange.com/questions/238997/prove-the-divergence-of-the-sequence-sinn-n-1-infty – Alex May 25 '14 at 10:05
  • I think that in this case (I can't say about other cases) the hint is as simple and direct as possibly wanted, and only requires substitution and some rather elementary trigonometric identity. +1 – DonAntonio May 25 '14 at 11:12
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No: By defination of derivatives;

$$f'(0)=lim_{h\to 0}\dfrac{f(h)-f(0)}{h-0}$$ $$=lim_{h\to 0}\dfrac{f(h)}{h}$$ $$=lim_{h\to 0}\dfrac{hsin(1/h)}{h}=lim_{h\to 0}sin(\dfrac 1 h)$$

It is enough to show that $lim_{x\to \infty}sin(x)$ does not exist.

if $x_n=\pi n$ for $n\in \mathbb N$, $lim_{n\to \infty}sin(x_n)=0$

if $x_n=(\pi/2)(4n-1)$, $lim_{n\to \infty}sin(x_n)=1$

Thus, $lim_{x\to \infty}sin(x)$ does not exist.

mesel
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