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How would I prove that the function

$f: \Bbb R \to \Bbb R$, where $f(x) = \begin{cases} x^2\cos(1/x) & x \neq 0 \\ 0 & x=0 \end{cases}$

is differentiable?

So far I have tried using the product rule, but gotten stuck with differentiating $\cos(1/x)$ from the definition of differentiation.

Martin Sleziak
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user284408
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  • You use the product rule for $x\not=0$. For $x=0$, use the definition of derivatives. –  Nov 29 '15 at 17:01
  • This question is one of the most duplicated here, as Google search for x^2 sin(1/x) differentiable would tell you. –  Nov 29 '15 at 18:22

1 Answers1

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Only point of concern is with $x=0$:

$$\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\dfrac{x^2\cos{\frac{1}{x}}}{x}=\lim_{x\to 0}\left(x\cos{\dfrac{1}{x}}\right).$$

Now as $\cos{\dfrac{1}{x}}$ is bounded in a neighborhood of the origin (besides $0$) we have $x\cos{\dfrac{1}{x}} \to 0$ as $x\to 0$

Learnmore
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  • How do you get the zero from xcos(1/x)? – user284408 Nov 29 '15 at 17:13
  • suppose $|\cos {\frac{1}{x}}|<M$(say) then what happens if you consider $x.M$ where $x\to 0$ – Learnmore Nov 29 '15 at 17:18
  • Okay, so how about when x doesnt equal zero? I am stuck with this bit too – user284408 Nov 29 '15 at 17:24
  • When it doesn't equal 0 it doesn't work. If your function oscillates within a bound (i.e. doesn't reach infinite values), you can be sure that it will be 0 by the sandwich theorem; but since it oscillates and you don't know exactly which value it takes, multiplying it by another number different than 0 doesn't assure anything about the product's value. – Fede Poncio Nov 29 '15 at 18:29
  • Why the down vote @jack – Learnmore Nov 30 '15 at 02:36
  • @Amartya: I didn't do the downvote. You can see my name under your post since I edited it. –  Nov 30 '15 at 02:38