Is $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$?

Solution 1:

A qualified "yes": If $f$ is continuous at $x_{0}$, and if $\lim\limits_{x \to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.

Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)

The claim follows from the Mean Value Theorem: If $\delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < \delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that $$ \frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t). $$ Since $|t - x_{0}| < |h|$, taking the limit as $h \to 0$ forces $t - x_{0} \to 0$, as well, so $$ f'(x_{0}) = \lim_{h \to 0} \frac{f(x_{0} + h) - f(x_{0})}{h} = \lim_{t \to x_{0}} f'(t). $$

(Continuity of $f$ was needed to invoke the Mean Value Theorem.)

Solution 2:

An easier proof, for the chosen answer, is by using the L'Hopital's rule. We know that $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Now if f is continuous at $a$ the we have a $\frac{0}{0}$ situation, and we can apply the L'Hopital's rule to see that if the limit of $f(x)$ when $x\mapsto a$ exists then it is equal to $f'(a)$.