I want to show the following:

Let $f: \mathbb{R} \to \mathbb{R}$ be continuous.

a) If $f$ is differentiable and for $x \ne 0$ the limit $\lim_{x\to 0} f'(x) = A$ exists, then $f$ is differentiable at $x = 0$ and $f'(0) = A$.

b) Show that the inverse is false, i.e. there exists a function $f$ which is differentiable at $x = 0$, but $\lim_{x\to 0} f'(x)$ does not exists.

For a), I worked out a proof, but I am unsure if the limit manipulations I used are okay and rigorous enough, so please can you comment on my solution and point out possible flaws?

Ok, in a) I have to show, that if the limit exists, then it obeys the continuity condition at $x = 0$, i.e. $$ \lim_{x\to 0} f'(x) = f'(0) $$ so I calculate (by using the limit definition of the derivative) \begin{align*} \lim_{x\to 0} f'(x) & = \lim_{x \to 0} \left[ \lim_{h \to 0} \left( \frac{f(x+h)-f(x)}{h} \right) \right] \\ & = \lim_{h \to 0} \left[ \lim_{x \to 0} \left( \frac{f(x+h)-f(x)}{h} \right) \right] \\ & = \lim_{h \to 0} \left[ \frac{1}{h} \lim_{x \to 0} \left( f(x+h)-f(x) \right) \right] \\ & = \lim_{h \to 0} \left[ \frac{1}{h} ( f(h) - f(0) ) \right] \\ & = f'(0) \end{align*} Thats my proof, in the last step I used the continuity of $f$ and the manipulations are possible, I think, because all limits exists. I never saw such manipulations, most proofs in my textbooks use $\epsilon/\delta$-Arguments, so I am unsure as how valid are such limit-exchange operations.

For b) the function $f(x) := |x|$ is differentiable at $x = 0$ but it's derivative is not continuous at $x = 0$.

  • 1
    $\begingroup$ In general, you cannot swap limits like this. To see why, compare $\lim_{n\to\infty}\lim_{m\to\infty}\frac{n}{m+n}$ and $\lim_{m\to\infty}\lim_{n\to\infty}\frac{n}{m+n}$. $\endgroup$ – Ayman Hourieh Feb 4 '13 at 21:52
  • 1
    $\begingroup$ Also, for b), your function isn't differentiable at $x=0$. $\endgroup$ – Ayman Hourieh Feb 4 '13 at 21:54
  • $\begingroup$ Ok, does this make my proof invalid? Are there circumstences when such limit swaping is valid? For b), what are better examples, to my mind I can just think of functions that have "breaks" at points like the $|\cdot|$ function? $\endgroup$ – StefanH Feb 4 '13 at 21:57
  • $\begingroup$ See the function in this question. $\endgroup$ – Brian M. Scott Feb 4 '13 at 22:08
  • $\begingroup$ For part a), you could use the Mean Value Theorem. See this post for details (it handles the one-sided case, but can be suitably modified for your problem). $\endgroup$ – David Mitra Feb 4 '13 at 22:08

As I said in the comments, swapping limits doesn't work in general. Let's prove a) using an $\epsilon-\delta$ argument.

Fix $\epsilon > 0$, and pick $\delta > 0$ so that: $$ |x| < \delta \implies |f'(x) - A| < \epsilon \tag{1} $$

This is always possible since $\lim_{x \to 0} f'(x) = A$.

Let $h \in (0, \delta)$. Since $f$ is continuous on $[0, h]$ and differentiable on $(0, h)$, by the mean value theorem we can find $\xi \in (0, h)$ so that:

$$ \frac{f(h) - f(0)}{h} = f'(\xi) $$

But $\xi$ satisfies (1), thus:

$$ \left| \frac{f(h) - f(0)}{h} - A\right| < \epsilon $$

Since our choice of $\epsilon$ is arbitrary, we conclude that $f$ is right differentiable at $x = 0$. A similar argument works for the left derivative, and the desired result follows.

For b), the standard example is $f(x) = \begin{cases} x^2 \cos(1/x) & x \ne 0 \\ 0 & x = 0 \end{cases}$. Try to prove that it's differentiable at $x = 0$ but the derivative isn't continuous there.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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