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 !

  • 6
    $\begingroup$ A limit that is $+\infty$ is very much like a limit that exists. Apply Lagrange's theorem and take limits $\frac{f(x)-f(0)}{x}=f'(c_x)$ for some $c_x\in(0,x)$. When you take limit when $x\to 0$, the left tends to $f'(0)$ and the right will have to tend to $+\infty$. $\endgroup$
    – plop
    Commented Jun 9, 2021 at 0:22
  • 5
    $\begingroup$ A theorem quite related to the things that you are asking is Darboux's theorem, that the derivative takes all values between any two. So, you can imagine that if the derivative is taking the value $f'(0)$ at $x=0$ and very large values at nearby $x$, then in between $0$ and $x$ it should also take all intermediate values. So, it will have to oscillate. $\endgroup$
    – plop
    Commented Jun 9, 2021 at 0:29

1 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.


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