# change order of limit

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

• In your example, $f'(0)=0$. Is the limit of the derivative zero? In general though, derivatives may not be continuous. Nov 30, 2019 at 5:20
• You have your equality if the limit on the right hand side exist (assuming that $f$ is differentiable everywhere). See here. Nov 30, 2019 at 5:22
• 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. Nov 30, 2019 at 6:43
• 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.
– user65203
Nov 30, 2019 at 8:27
• Nov 30, 2019 at 9:40

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. $$f(x)=|x|$$.