Assume this function has a value $0$ at $x=0$.

If the right hand derivative of a derivable function is equal to the right hand limit of the derivative function (same for the left), why aren't functions like the one above continuously differentiable?

For a function to be continuous, limits should be finite and equal to value at that particular point.

For this function derivative using first principle yields $0$.

The RHD, LHD is $0$, I think. So why isn't the derivative continuous?

For finding the RHD and LHD, I have seen answers resorting to using known derivatives. My question is: why don't we use the first principle itself? Why do we get different answers then?


1 Answer 1


The function, let's call it $f$, is indeed differentiable. For $x\ne0$ we have $$ f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x} $$ using the chain and product rules.

At zero, using the definition of derivative, $$ f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}= \lim_{h\to0}h\sin\frac{1}{h}=0 $$

On the other hand the limit $\lim\limits_{x\to0}f'(x)$ does not exist. If we compute it on the sequence $a_n=\frac{1}{2\pi n}$, we have $$ \lim_{n\to\infty}f'(a_n)=\lim_{n\to\infty} \left(\frac{1}{2\pi n}\sin(2\pi n)-\cos(2\pi n)\right)=-1 $$ whereas, over $b_n=\frac{1}{\pi+2\pi n}$ we have $$ \lim_{n\to\infty}f'(b_n)=\lim_{n\to\infty} \left(\frac{1}{\pi+2\pi n}\sin(\pi+2\pi n)-\cos(\pi+2\pi n)\right)=1 $$ Therefore $f'$ is not continuous at $0$.

Why? Because it's so. A function needn't be continuous, even if it is the derivative of another function.

  • $\begingroup$ I accept this,but as i mentioned in my question,why cant i calculate the right and left limits of the derivative using the definitions of RHD and LHD .They both come out to be 0. $\endgroup$ Mar 31, 2017 at 17:36
  • $\begingroup$ @MarzooqAbdulKareem No, the left-hand and right-hand limits of the derivative are not zero. The examples I gave are good for the RHD, but they can easily be adapted for the LHD. $\endgroup$
    – egreg
    Mar 31, 2017 at 17:41
  • $\begingroup$ Yes,agreed.But what i am asking is that if RHD exists and it is differentiable,why do i need to calculate the right hand limit itself.Cant i directly use the value of RHD. Basically ,i dont understand why if a function is differentiable,its derivative need not be continuous? Graphically it might sort out somehow,but somehow just cant seem to digest it. $\endgroup$ Mar 31, 2017 at 17:54
  • $\begingroup$ @MarzooqAbdulKareem The derivative at $0$ exists, no doubt about that. However the derivative, as a function, is not continuous at $0$. $\endgroup$
    – egreg
    Mar 31, 2017 at 22:06

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.