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Do we compute $\lim_{x\to a}{f'(x)}$ whenever we are computing $f'(a)$? I got to this question like this :

$$f'(x)=\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\to a}{\frac{f'(x)}{1}}= \lim_{x\to a}{f'(x)}$$ But if this is true, then why the derivative is indeterminate when $f$ is discontinuous in $x=a$ but still has limit in that point?

Arya369
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1 Answers1

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Fix $a\in\mathbb R$. You have that $f'(a)$ may exist, whereas $\lim_{x\to a}f'(x)$ may not. A classical example of such function is$$f(x)=\begin{cases} (x-a)^2\sin\left(\frac{1}{x-a}\right)&x\neq a\\ 0&x=a. \end{cases}$$ However, if $\lim_{x\to a}f'(x)$ exist, then $f'(a)$ exist and $$\lim_{x\to a}f'(x)=f'(a).$$

Surb
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  • For the people wondering about the proof of the last statement https://math.stackexchange.com/questions/857139/derivative-of-continuous-function-exists-if-limit-of-derivative-exists – Severin Schraven Jan 25 '25 at 14:11