Can there exist a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $\lim\limits_{x \to 0}f'(x) = + \infty$.

Question

I think that the following proposition is true:

There exists a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(0) = 0$ and for which there exists a sequence $(x_n)_n$ in $\mathbb{R}$ such that $x_n \rightarrow 0$ and $f'(x_n) \rightarrow +\infty$.

I came up with the following function: $$f: \mathbb{R} \rightarrow \mathbb{R}:x \rightarrow f(x ) =\begin{cases} x^2 \sin{(1/x^2)} &\mbox{if }x\neq 0,\\ 0 &\mbox{if } x = 0.\\ \end{cases}$$ It's fairly simple to prove that $f$ is differentiable (on its domain) with its derivative being$$f': \mathbb{R} \rightarrow \mathbb{R}:x \rightarrow f'(x) =\begin{cases} 2x \sin{(1/x^2)} - \frac{1}{x} \cos{(1/x^2)} &\mbox{if }x\neq 0,\\ 0 &\mbox{if } x = 0.\\ \end{cases}$$ Then the sequence $(x_n)_n$ given by $x_n = \frac{1}{\sqrt{\pi + 2n\pi}}$ will converge to zero and $f'(x_n) = \sqrt{\pi + 2n\pi}$ will go to $+\infty$.

Now I was wondering if the following proposition is true:

There exists a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\lim\limits_{x \to 0}f'(x) = + \infty$.

Although $\lim\limits_{n \to \infty}f'(x_n) = + \infty$, my example does not fullfil the condition of $\lim\limits_{x \to 0}f'(x) = + \infty$ (the function oscilates between $-\infty$ and $+ \infty$). I do not know quite sure if the proposition is true or not, but I could not come up with any examples. I also tried proving it with L'Hôpital's rule and with the Mean value theorem, but to no avail. I feel like there is a very simple example that I am overseeing. Any help is welcome !

Answer 1

NO. Suppose $f(x)$ is continous at $x=0$ and is differentiable for $x>0$. Suppose $f'(x)\to\infty$ as $x\to 0^+.$

Let $(x_k)_{k\in \Bbb N}$ be a positive sequence converging to $0,$ such that $\forall y\in (0,x_k]\;(f'(y)\ge k).$

For any $z\in (0,x_k)$ there exists $y\in (z,x_k)$ with $f(x_k)-f(z)=(x_k-z)f'(y)\ge (x_k-z)k.$ Now let $z\to 0,$ so $$f(x_k)-f(0)=\lim_{z\to 0}f(x_k)-f(z)\ge \lim_{z\to 0} \,(x_k-z)k=x_k\cdot k.$$ Hence we have $$(\bullet)\quad \frac {f(x_k)-f(0)}{x_k-0}\ge k.$$ Now $x_k\to 0$ as $k\to\infty$ so $(\bullet)$ shows that $f'(0)$ does not exist.