# Limit of the ratio of a nowhere differentiable function to a polynomial

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.

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.

• 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].) Commented Jun 29, 2021 at 16:10
• 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.
– zhw.
Commented Jun 29, 2021 at 19:49