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For function $f(x)$, let \begin{equation} f(c)=\lim_{x \to c} f(x). \end{equation} Under what conditions the following holds? \begin{equation} f'(c)=\lim_{x\to c} f'(x). \end{equation} For instance, let $f(x)=\frac{\sin(x)}{x}$, \begin{equation} f(0)=\lim_{x\to 0}\frac{\sin(x)}{x}=1. \end{equation} Is the following true? \begin{equation} f'(0)=\lim_{x\to 0}\left(\frac{\sin(x)}{x}\right)' \end{equation}

user
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jsmath
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    In your example, $f'(0)=0$. Is the limit of the derivative zero? In general though, derivatives may not be continuous. – Angina Seng Nov 30 '19 at 05:20
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    You have your equality if the limit on the right hand side exist (assuming that $f$ is differentiable everywhere). See here. – Arctic Char Nov 30 '19 at 05:22
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    It is true if $f$ is continuously differentiable at $x=c$. I'm not sure if that is an answer that is satisfactory for you though. – Tucker Nov 30 '19 at 06:43
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    For any function $g$, for

    $\lim_{x\to c}g(x)=g(c)$ to hold, it suffices that $g$ be continuous at $c$.

    So your question can be rephrased as "under what conditions is a derivative continuous?". A sufficient condition is the function twice differentiable, but this is not necessary.

     –  Nov 30 '19 at 08:27
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    Related: https://math.stackexchange.com/questions/257907/prove-that-fa-lim-x-rightarrow-afx – Hans Lundmark Nov 30 '19 at 09:40
  • @ArcticChar: But $f(x)=\frac{sin(x)}{x}$ is not differentiable at $x=0$. – jsmath Dec 01 '19 at 17:02
  • @JohnSmith Your $f$ is differentiable at $x=0$. – Arctic Char Dec 01 '19 at 20:30

1 Answers1

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Recall that

$$f(c)=\lim_{x \to c} f(x)$$

is true if and only if $f(x)$ is continuous at $x=c$ and therefore

$$f'(c)=\lim_{x\to c} f'(x)$$

is true if and only if also $f'(x)$ is continuous at $x=c$.

For the example

$$f(x)=\frac{\sin(x)}{x} \implies f'(x)=\frac{x\cos x-\sin x}{x^2}$$

by $f(0)=1$ and $f'(0)=0$, are continuous at $x=0$ and both identities hold.

But the property is not true in general, e.g. $f(x)=|x|$.

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