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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?

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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.

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  • $\begingroup$ 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

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