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

Question

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.

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.