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So the question goes as following:

Let $f(x)=x^2\sin(1/x)$ if $x\not=0$

show that $f'(x)$ exist at every $x$ but $f'$ is not continous at $x=0$.

I'm not really sure how to show it.

Usually I would show that when x approaches 0 from either side it will have different limits, but this is clearly not the case here.

All help is welcomed!

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  • $\begingroup$ in order for a function to be characterized discontinuous at some point $x_0$, this point must be in the domain of the function. If so, what is the value of your function $f(x)$ at $x_0=0$ ? $\endgroup$
    – KonKan
    Commented Dec 3, 2017 at 21:20
  • $\begingroup$ The discontinuity at $x=0$ is removable, so the function is extended to $f(0)=0$. The derivative at $0$ is $0$. It remains to check that $\lim_{x\to 0} f'(x)$ either does not exist, or does exist but is not $0$. (hint: it's the former) $\endgroup$
    – ziggurism
    Commented Dec 3, 2017 at 21:23
  • $\begingroup$ Right. But do I have to define f(0)=0? if it does not exist? $\endgroup$ Commented Dec 3, 2017 at 21:25
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    $\begingroup$ Yes, you have to define $f(0)=0.$ If the problem didn't do this for you, it is not correctly posed. $\endgroup$
    – ziggurism
    Commented Dec 3, 2017 at 21:25
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    $\begingroup$ factor the derivative? make it differentiable? No one asked you to find the derivative of the derivative (aka second derivative). The question before you is whether this function $f'(x)$ is continuous. What value does $\cos(\frac{1}{x})$ approach as $x$ approaches 0? $\endgroup$
    – ziggurism
    Commented Dec 3, 2017 at 21:30

3 Answers 3

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The correct statement goes something like this:

The function $$ f(x)=\left\{ \begin{array}{c} x^2\sin(\frac{1}{x}), \ \ x\neq 0 \\ 0, \ \ x=0 \end{array} \right. $$ is differentiable everywhere and (thus) continuous everywhere.

Its derivative (use product + chain rule of differentiation for $x\neq 0$ and apply the definition of the derivative at $x=0$) is: $$ f'(x)=\left\{ \begin{array}{c} 2x \sin(\frac{1}{x})-\cos(\frac{1}{x}), \ \ x\neq 0 \\ 0, \ \ x=0 \end{array} \right. $$ which is discontinuous at $x=0$. (the limit of the upper branch for $x\rightarrow 0$ does not exist).

P.S.: Here is a quick argument to show why $$\lim_{x\rightarrow 0}f'(x)=\lim_{x\rightarrow 0}\big(2x \sin(\frac{1}{x})-\cos(\frac{1}{x})\big)$$ does not exist: Let us suppose that the limit exists and is a real number $l$: $$\lim_{x\rightarrow 0}f'(x)=\lim_{x\rightarrow 0}\big(2x \sin(\frac{1}{x})-\cos(\frac{1}{x})\big)=l\in\mathbb{R}$$ Thus: $$ \cos(\frac{1}{x})=2x \sin(\frac{1}{x})-f'(x), \ \ \ \ x\neq 0 $$ and since $\lim_{x\rightarrow 0}x \sin(\frac{1}{x}\big)=0$ (remember that: $0\times bounded = 0$), after taking limits at $x\rightarrow 0$ of both sides of the last equation we arrive at: $$ \lim_{x\rightarrow 0}\cos(\frac{1}{x})=-l\in\mathbb{R} $$ which is clearly absurd, since it is well known (and almost evident) that the $\lim_{x\rightarrow 0}\cos(\frac{1}{x})$ does not exist.

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  • $\begingroup$ right. However in the formula in my book they explain it as following: $f'(x)=lim_{h->0}\frac{f(h)}{h} =lim_{h->0}hsin(\frac{1}{h})=0$ I'm not quite sure what they did. They redfined the derivative for the function? $\endgroup$ Commented Dec 3, 2017 at 21:37
  • $\begingroup$ They probably mean that: $f'(x)=lim_{h->0}\frac{f(h)-f(0)}{h-0} =lim_{h->0}\frac{f(h)}{h}$. (remeber that by definition: $f(0)=0$). $\endgroup$
    – KonKan
    Commented Dec 3, 2017 at 21:40
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    $\begingroup$ I think that I get it now. Your latest edit was very helpful, thank you! $\endgroup$ Commented Dec 3, 2017 at 21:58
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hint

For $x\ne 0$,

$$f'(x)=2x\sin (1/x)-\cos (1/x) $$

$\lim_0 2x\sin (1/x)=0$

$\lim_0 \cos (1/x) $ doesn't exist.

so $\lim_0 f'(x)$ doesn't exist.

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There are several helpful comments and answers already, but I want to point out this line:

Usually I would use show that when $x$ approaches $0$ from either side it will have different limits, but this is clearly not the case here.

Beware of pronouns! What you mean by it here? Based on what you write next, it sounds like you're looking at $$ \lim_{x\to 0} x^2 \sin(1/x) = \lim_{x\to 0} f(x) $$ You are right that his limit does exist (it equals zero). But that just tells you the function is continuous at $0$.

You want to show $f'(x)$ is not continuous at $0$. That means you are looking at the limit $$ \lim_{x\to 0} f'(x) $$ Once you calculate what $f'(x)$ is away from zero, you will see why the limit above does not exist.

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