Darboux's theorem says that $f'$ has intermediate value property. More precisely,
( Darboux's theorem) If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then $\exists$ c in (a,b) such that $f'(r)=c$.
Thus Darboux's theorem implies that $f'$ cannot have any simple discontinuities on $[a,b].$
I have a question as follows:
Is there a function $f$ satisfying $\displaystyle \lim_{x\to d} f'(x)=\infty$ for some $d \in (a,b)$ under the assumptions in Darboux's theorem?
If so, I can conclude that for each $x \in (a,b),$ either $(i)$ $f'$ is continuous at $x$ or $(ii)$ $f'$ oscillates near $x.$
It seems that there is no differentiable function $f$ on $[a,b]$ satisfying function $\displaystyle \lim_{x\to d} f'(x)=\infty$ and $\exists f'(d)$ for some $d \in (a,b)$.
Please let me know if you have any idea or comment for my question. Thanks in advance!
Best Answer
If $f$ is differentiable on $(a,b)$ and $x_n \to d \in (a,b)$ from the right, we have
$$f'(d) = \lim_{n \to \infty} \frac{f(x_n) -f(d)}{x_n - d}$$
By the MVT there is a sequence $\xi_n \in (d,x_n)$ such that
$$f'(d) = \lim_{n \to \infty} \frac{f(x_n) -f(d)}{x_n - d} = \lim_{n \to \infty} f'(\xi_n)$$
However, if $f'(x) \to \infty$ as $x \to d$, then we must have $f'(\xi_n) \to \infty$ since $\xi_n \to d$, a contradiction.
Hence, your function does not exist.