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

(a) Does there exist a function $f$ defined on the open interval $(a,b)$ such that $f'(b^-)$ exists, and $\lim_{x\to b-}f'(x)\neq f'(b^-)$, or (b) where $f'(b^-)$ exists and $\lim_{x\to b-}f'(x)$ does not exist?

Since $f(b)$ is undefined, define $$f'(b^-)=\lim_{h\to0+}\frac{f(b-h)-f(b-2h)}h.$$ Are there any difficulties with this definition as compared to the standard definition of the one-sided derivative?

Just reading my analysis textbook and thought this would make an interesting problem.

Related: $f(x)=x^2\sin\frac1x$ has a derivative which is defined at $0$ (equal to $0$), but $\lim_{x\to 0}f'(x)$ does not exist. (c) Is it always true that if the limit exists, it is equal to $f'(0)$? Even more curiously, $\limsup_{x\to0}f'(x)+\liminf_{x\to0}f'(x)=2f'(0)$ for this function. (d) Is this always the case, when the quantity on the left side of the equality is defined?

share|cite|improve this question
This post may be of interest. – David Mitra Dec 17 '12 at 21:17
$f(x)=\sin(2\pi \log_2|x|)$ has $f'(0^-)=0$ but $\lim_{x\to 0^-}f'(0)$ doesn't exist, so (b) has a positive answer. – user108903 Dec 17 '12 at 21:23
up vote 2 down vote accepted

Consider e.g. a function of the form $f(x) = g(\log_2(b-x))$ where $g$ is periodic with period $1$. This satisfies $f(b-2x) = f(b-x)$, and thus according to your definition $f'(b-) = 0$, while $f'(x) = - \dfrac{g'(\log_2(b-x))}{(b-x) \ln 2}$, so $\lim_{x \to b-} f'(x)$ will not exist unless $g$ is constant.

EDIT: On the other hand, if $f$ is differentiable on $[b-2h,b-h]$ the Mean Value Theorem says
$ \dfrac{f(b-h) - f(b-2h)}{h} = f'(\xi)$ for some $\xi \in [b-2h,b-h]$, so if $\lim_{x \to b-} f'(x)$ exists then so does $f'(b-)$ and the two are equal.

Moreover, if $f'(b)$ exists, since $$\frac{f(b-h) - f(b-2h)}{h} = \frac{f(b-h) - f(b)}{h} - 2 \frac{f(b-2h) - f(b)}{2h}$$ we must have $f'(b-) = f'(b)$.

EDIT: As in your comments, suppose $g$ and $h$ are continuous and decreasing on $[0,\epsilon]$ with $h(x) < g(x) < b$ for $x > 0$ and $g(0) = h(0) = b$, and you define $f'(b-) = \lim_{x \to 0+} \dfrac{f(g(x)) - f(h(x))}{g(x) - h(x)}$.
Define a sequence $x_n$ by $x_0 = \epsilon$ and $h(x_{n+1}) = g(x_n)$. Then $x_{n+1} < x_n$ and $\lim_{n \to \infty} x_n = 0$. Take any nonconstant continuous function $f$ on $[h(x_0), g(x_0)] = [h(x_0), h(x_1)]$ with $f(h(x_0)) = f(h(x_1))$, and then define $f$ on $[h(x_n),h(x_{n+1})] = [g(x_{n-1},g(x_n)]$ iteratively by $f(t) = f(h(g^{-1}(t)))$. We obtain a continuous function $f$ on $[h(x_0), b)$ with $f(g(x)) = f(h(x))$, and thus $f'(b-) = 0$, but $\lim_{x \to b-} f(x)$, and thus also $\lim_{x \to b-} f'(x)$, does not exist.

share|cite|improve this answer
I hadn't thought about using $\log_2$ to get periodicity that way. Would you consider this an artifact of my formula, or would a more clever way of letting the points approach zero of the form $\lim_{x\to0+}\frac{f(g(x))-f(h(x))}{g(x)-h(x)}$ where $g,h\to b^-$ as $x\to0^+$ prevent that? – Mario Carneiro Dec 17 '12 at 21:33
I'm not sure what "that" you are hoping to prevent. – Robert Israel Dec 17 '12 at 21:37
Well, in your example, you have an oscillating function, and I sample the function at two points, but you've set up the function so that the two points are always a period apart, so I can't "tell" that the function is oscillating, and hence declare it to have zero derivative. Indeed, it is clear that your function is "adversarial", in the sense that the $\log_2$ part was chosen specifically so that this behavior occurs ($\log_3$ would not have worked for this). My question is whether I could define a $g$ and $h$ such that you can't hide periodicity in this manner. – Mario Carneiro Dec 17 '12 at 21:42
@MarioCarneiro $\frac{f(g(x))-f(h(x))}{g(x)-h(x)}=f'(\xi)$ with $\xi\in(h(x),g(x))$ will always hold by MVT. Then if $\lim f'$ exists, it will also equal this revised definition of $f(b^{-})$. – Hagen von Eitzen Dec 17 '12 at 21:47
@HagenvonEitzen In this case, my attempt is to make it so that whenever $\lim_{x\to b}f'(x)$ doesn't exist, the revised definition of $f(b^-)$ will not be defined either. (The example in my mind right now is $g(x)=b-x$ and $h(x)=b-x^2$.) – Mario Carneiro Dec 17 '12 at 21:52

(a) If $\lim_{x\to b} f'(x)$ and $f'(b^{-})$ exist, then the limit equals $f'(b^-)$ because $\frac{f(b-h)-f(b-2h)}h=f'(b-\theta h) $ with some $\theta\in(1,2)$.

You've already given $f(x)=x^2\sin\frac1x$ as example for (b). Indeed, we also have $f'(0^-)=0$.

For (c) the same MVT argument applies as for (a).

(d) Let $g$ be smooth periodic and $$f(x)=\begin{cases}x^2\cdot g\left(-\frac1x\right)&\text{if }x>0\\ 0&\text{if }x=0.\end{cases}$$ Then $f'(0)=0$ and $f'(x)=2x g(-1/x)+g'(1/x)$, hence $\limsup f'=\limsup g'$ and $\liminf f'=\liminf g'$. If $g$ is asymmetric, we will have $\limsup f'+\liminf f'\ne0$. Indeed, consider $g(t)=\sin t+\sin 2t$, a first approximation to a sawtooth curve, which has $g'(0)=3$ but $g'(t)>-3$ for all $t$.

share|cite|improve this answer

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.