I'm constructing a continuous function $f$ which is well defined on $\mathbb R$ and its derivative $g$ does not exists on a countable infinite points. This kind of functions is easy to find. For example, a continuous function consisting a series of connected line segments. I think the $g$ is actually continuous on its domain, if its domain is restricted on where it exists (correct me if I am wrong). Here its domain is a dense set. Then I am wondering if we can find an example or a counterexample which succeed or fail the L'Hospital rule, of which the numerator and denominator are continuous on $\mathbb R$, and their derivative exists almost everywhere.


I guess I should be more specific. I want the points where the derivative does not exist are concentrated around the point where L' Hospital will be used. For example, L' Hospital rule will be used at $x=0$ and the numerator and denominator's derivatives do not exist on $\{x=\frac{1}{n}\}$


I note that in common L'Hospital rule statement and proof, it is assumed that in $\lim\frac{f(x)}{g(x)}$, $f$ and $g$ are not only continuous on an interval (neighborhood of $x_0$, where the rule will be used), but also differentiable on the interval everywhere. See here. I am wondering if this condition is too strong. I think the derivatives of $f$ and $g$ are not necessary to exist everywhere on the interval but only need the $\lim \frac{f'}{g'}$ to exist, to apply L'Hospital rule. In other words, can we relax the condition by allowing $f'$ and $g'$ to exist outside a smaller, in some sense (e.g. measure zero ?), subset of the interval.

  • $\begingroup$ Re Update 2: As discussed below this ultimately depends on obtaining a monotonicity theorem of the form "$f$ has some property and $f'(x)>0$ almost everywhere, therefore $f$ is nondecreasing." For more than you might care to know about monotonicity theorems of this type see Chapter 11 of Andy Bruckner's monograph Differentiation of Real Functions. $\endgroup$ Dec 31, 2015 at 17:16

1 Answer 1


Maybe, perhaps, this is what you are thinking of? I will pose as another (related) problem.

Problem. Suppose $f$ is continuous in some neighborhood $(x_0-\delta,x_0+\delta)$ of a point $x_0$ and there is a "small" set $N$ so that $f'(x)$ exists for each $x\not\in N$ and moreover $$ \alpha < f'(x) < \beta$$ for each $x\in (x_0-\delta,x_0+\delta)\setminus N$.

Can we conclude that, for all $x\in (x_0-\delta,x_0+\delta) $ $$ \alpha \leq \frac{f(x)- f(x_0)}{x-x_0} \leq \beta?$$

If so then you could use this result for your L'Hopital's rule. For example if $f(x_0)=0$ and $g(x_0)=0$ then $$ \lim_{x\to x_0} \frac{f(x)}{g(x)} = \lim_{x\to x_o, x\not\in N} \frac{f'(x)}{g'(x)} $$ would be true under the right hypotheses. You need (i) $N$ to be suitably small (i.e., it answers the problem above) (ii) you need $f$ and $g$ differentiable outside $N$ and (iii) you need
both limits to exist: $$\lim_{x\to x_o, x\not\in N} {f'(x)} $$ and $$ \lim_{x\to x_o, x\not\in N} {g'(x)} \not= 0 $$

So really your fundamental problem here is a monotonicity problem. The answer is that for continuous $f$ and a countable set $N$ the answer to the problem posed is "yes". For continuous $f$ and a measure zero set $N$ the claim might be false. For absolutely continuous $f$ and a measure zero set the answer is again "yes."

I hope this isn't too brief or cryptic for your purposes.

  • $\begingroup$ Thanks for this original response. I am interested the second last paragraph in your answer. Could you refer me to some more detailed source about monotonicity problem? And I guess I could make my purpose more clear by updating the question. $\endgroup$
    – Hua
    Dec 31, 2015 at 5:19
  • $\begingroup$ Work on this problem (or ask for help here): If $f:[a,b]\to\mathbb R$ is continuous and $f'(x)>0$ for all $x$ with at most countably many exceptions then $f$ is nondecreasing. $\endgroup$ Dec 31, 2015 at 17:08

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