Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that this can not be true due to counter-examples but I don't know where the error in my reasoning is.

Assumption: If $f(x)$ is differentiable in $\mathbb{R}$ then the derivative $f'(x)$ is continuous in $\mathbb{R}$.

Faulty Proof: For every $c \in \mathbb{R}$, using the mean value theorem for $f(x),$ on the interval $x \in [c, c + h] $ where $h$ is positive.

$$ \frac{f(c + h) - f(c)}{h} = f'(\xi(h)) $$

Where $\xi(h) \in (c,c+h)$. Because this equation holds for every $h>0$. It must hold in the limit as $h \rightarrow 0^+$.

$$ \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h)) $$

But the left side of the equation is the right one sided derivative.

$$ f'_{+}(c) = \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h)) $$

The same can be done for $h$ being negative, but because of differentiability at every point the left and right derivatives must be equal.

$$ f'(c) = f'_{+}(c) = f'_{-}(c) = \lim_{h\to 0^-}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^-}f'(\xi(h)) $$

As $h \rightarrow 0^+$, $\xi(h) \rightarrow c$. So because the limit $\lim_{h\to 0^+}f'(\xi(h))$ exists and $\xi(h) \neq c$, it is equal to $\lim_{x\to c^+}f'(x)$

It follows that $\lim_{x\to c^+}f'(x) = \lim_{x\to c^-}f'(x) = f'(c)$ so the function $f'(x)$ is continuous.

share|cite|improve this question
You just proved that $\lim_{h\to 0}f'(\xi(h))=f'(c)$. How does it follow from this that $\lim_{x\to c}f'(x)=f'(c)$? For all you know, $\xi(h)$ could always be rational, for example. – Andrés E. Caicedo Jul 24 '14 at 14:49
The proof would be correct for showing that if the limit of the derivatives exists, that limit is the derivative at $c$. However, as tracing a standard counterexample will show, the limit need not exist. – André Nicolas Jul 24 '14 at 14:52
Have you try plugging the counterexample into $f$ in the above proof to see where it went wrong? – Gina Jul 24 '14 at 14:54
up vote 10 down vote accepted

The issue is that you can't compute limits along particular paths like $\xi(h)$.

If you prove that $\lim_{n \to +\infty} f(p_n)$ exists for some $p_n \to x_0$, you cannot deduce that $\lim_{x \to x_0} f(x)$ exists.

share|cite|improve this answer
$\xi$ may not even describe a path, since it could fail to be continuous. – Andrés E. Caicedo Jul 24 '14 at 14:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.