My professor said that the title statement might not always be the case and gave $$x^2 \sin\left(\frac{1}{x}\right)$$ at $x=0$ as a counter-example.

But I don't seem to understand its differentiability and continuity at the point $x=0$.

Any explanation and if possible a better example are highly appreciated.

  • 2
    $\begingroup$ Maybe helpful: math.stackexchange.com/questions/2776837/…. Your function (assuming you define it as $0$ at $x=0$) provides us an example of a function being differentiable at a point without the derivative being continuous there. $\endgroup$ – Minus One-Twelfth May 24 at 11:19
  • 2
    $\begingroup$ This is an example of a function that is differentiable; but whom’s derivative is not continuous... So the title should be the other way around $\endgroup$ – Maximilian Janisch May 24 at 11:40

Since $x = 0$ is not in the domain of $f(x)$, let's define $f(0) = 0$. Here's how it becomes continuous if we do so.

Continuity at $x = 0$:

$$\lim_{x \to 0} x^2 \sin(\frac{1}{x}) = 0$$

The function is continuous because $\sin(t) \in [-1,1] \quad \forall t$.

Differentiability at $x = 0$:

$$\lim_{h \to 0} \frac{h^2 \sin(\frac{1}{h}) - 0}{h}$$ $$=\lim_{h \to 0} h \sin(\frac{1}{h}) = 0$$

This will hold for the LHD as well. I have only considered the RHD. (Verify it!)

Therefore $f(x)$ is differentiable as well.

Now the above explanation only holds if we define $f(0) = 0$ and not otherwise.

Now coming to the statement in the title of the question, it doesn't make much sense. Because if the derivative of a function is continuous, it is implicit that the derivative exists at that point.

Edit: In this case the derivative is not continuous. For this consider $f'(x)$ at $x \neq 0$ (can be simply obtained by applying the product rule and chain rule to $f(x)$), with the fact that $f'(0) = 0$ and observe that it indeed isn't continuous at $x =0$. Maybe your professor meant that just because the derivative exists doesn't imply that it is continuous.


I think your professor gave you that

$f(x)=x^2 \sin(\frac{1}{x})$, if $x \ne 0$ and $f(0)=0.$

For $x \ne 0$ we have $f'(x)=2x \sin(\frac{1}{x})-\cos(\frac{1}{x})$ and $f'(0)=0.$

Now let $x_n= \frac{1}{n \pi}.$ Then $x_n \to 0$, but $f'(x_n)=(-1)^{n+1}.$

Hence $(f'(x_n))$ does not converge (t0 $f'(0)$). This shows that $f'$ is not continuous at $x=0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.