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Should $\lim_{t\to x} f'(t)$ need to exist for function $f$ to be differentiable at $x$?

Suppose the function $f$ is differentiable at $x$. This means $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = L$$ exists.

Now is it necessary that $ \lim_{t \to x} f'(t)$ exists?

If it should exist, then this is not happening for example below.

$$f(x) = \begin{cases} x^2\sin(1/x),&x>0,\\ 0,&x\leq 0 \end{cases} $$

This function is differentiable at $0$, i.e. $f'(0) = 0$. But, $\lim_{t\to0} f'(t)$ doesn't exist. As $$f'(t) = 2t\sin(1/t) - \cos(1/t)$$

Any reasoning behind this?

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    $\begingroup$ You've provided a counterexample and shown adequately that it is a counterexample. I'd say there is reasoning behind this and you have provided it, so I'm not sure what your question is. $\endgroup$ Jul 30 '17 at 6:32
  • $\begingroup$ My question is: shouldn't this limit exist? Well, I have counter example,but I am not getting why this is happening. $\endgroup$
    – MeetR
    Jul 30 '17 at 6:37
  • $\begingroup$ @spaceisdarkgreen any other counter example ? So that I can understand it better. $\endgroup$
    – MeetR
    Jul 30 '17 at 7:43
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    $\begingroup$ I would suggest just thinking hard about what's going on in this example. The function oscillates increasingly rapidly on approach to the origin to the point that its derivative jumps rapidly between zero and one (and you can come up with modifications where the derivative is unbounded). And yet the function is squeezed horizontally through the $\pm x^2$ envelope at the origin so its derivative must vanish at that point. $\endgroup$ Jul 30 '17 at 18:55
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If $f$ is differentiable it is continuous. Of course, the derivative need not be continuous.

Note that

$$\lim_{t \to x} \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} = \lim_{t \to x} f'(t)$$

need not exist nor equal

$$\lim_{h \to 0} \lim_{t \to x} \frac{f(t+h) - f(t)}{h} = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = f'(x) $$

Read about double sequences and double limits in a real analysis book. The truth lies there.

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Let $f(x) = x^2\cdot \chi_{\mathbb Q}(x).$ Then $f'(0)=0,$ but $f'(x)$ fails to exist if $x\ne 0.$ So of course there is no hope of $f'(0)$ being any kind of limit of $f'(x)$ as $x\to 0$ through nonzero values.

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  • $\begingroup$ Great example.. Thanks. $\endgroup$
    – MeetR
    Jul 31 '17 at 7:49
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On the other hand, if $\lim_{x\to a-} f'(x) = \lim_{x\to a+} f'(x) =L$, then also $f'(a) = L $.

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  • $\begingroup$ In example given in description, this limit doesn't exist. That's my question, for function to be derivative at that point, should this limit exist? $\endgroup$
    – MeetR
    Jul 30 '17 at 7:29
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    $\begingroup$ This is false, unless you assume that $f$ is continuous at $a$. $\endgroup$
    – egreg
    Jul 30 '17 at 7:49
  • $\begingroup$ @egreg Thanks.This is I wanted to know. $\endgroup$
    – MeetR
    Jul 30 '17 at 7:59
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    $\begingroup$ @MeetR But the limit of the derivative at $a$ need not exist even if the function is differentiable in a neighborhood of $a$. $\endgroup$
    – egreg
    Jul 30 '17 at 8:01
  • $\begingroup$ @egreg. That's right. I forgot that condition (continuity). $\endgroup$
    – md2perpe
    Jul 30 '17 at 11:08

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