Let $f(x)$ be a continuous function that—

  • maps the closed interval [0, 1] to [0, 1],
  • equals 0 at 0,
  • does not equal 0 anywhere except at 0, and
  • is nowhere differentiable on its domain.

Let $g(x)$ be a polynomial that—

  • maps the closed interval [0, 1] to [0, 1],
  • bounds $f$ from above, and
  • equals 0 at 0.

My question is: Does the limit $\lim_{x\to 0^+} f(x)/g(x)$ exist? If not, what are the weakest conditions required on $f$ for the limit to exist?

I know that by L'Hôpital's rule, the limit exists when $f(x)$ is differentiable on some interval $(0, \epsilon)$, but I don't know whether the limit still exists in this case when $f$ is not required to be differentiable. This question is neither homework nor a self-study assignment, nor is this coursework.


1 Answer 1


Let $w(x):[0,1]\to [1/2,1]$ be continuous and nowhere differentiable. Define

$$v(x)=\frac{x}{2}+\frac{x}{4}\sin (\ln x),$$

for $x\in (0,1],$ with $v(0)=0.$ Then $v$ is differentiable on $(0,1]$ and $v$ is continuous on $[0,1].$ Note that $x/4\le v(x)\le 3x/4.$

Set $f(x) = v(x)w(x)$ and $g(x)=x.$ Then $f,g$ satisfy the given conditions. The only condition that's not immediately clear is that $f$ is nowhere differentiable. To verify it, let $x\in (0,1]$ and suppose $f'(x)$ exists. Then $(f/v)'(x)$ exists, i.e., $w'(x)$ exists. That's a contradiction. To check $x=0,$ observe

$$\frac{f(x)}{x} = w(x)\left(\frac{1}{2}+\frac{1}{4}\sin (\ln x)\right).$$

As $x\to 0^+,$ $w(x)\to w(0)\ne 0,$ and the factor on the right oscillates between $1/4$ and $3/4.$ This oscillation shows $f'(0)$ fails to exist. Thus $f$ is nowhere differentiable, and of course this also shows $\lim_{x\to 0^+} f(x)/g(x)$ fails to exist.

  • $\begingroup$ Thank you for the example. Can you find an example where the limit does not exist even if $f$ is Lipschitz continuous? (I know a Lipschitz continuous function is differentiable almost everywhere, but not whether it can be differentiable except on a dense set of measure zero or whether $f$ is differentiable on (0, $\epsilon$) whenever it is Lipschitz continuous on [0, 1].) $\endgroup$
    – Peter O.
    Jun 29, 2021 at 16:10
  • $\begingroup$ Take $f(x)=v(x)$ where $v(x)$ is as above, and $g(x)=x.$ Check that $v'(x)$ is bounded, hence $v$ is Lipschitz. $\endgroup$
    – zhw.
    Jun 29, 2021 at 19:49

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