0
$\begingroup$

Assume $f:[x_0,x_0+\epsilon)\to\mathbb{R}$ is a continuous function which is differentiable in $(x_0,x_0+\epsilon)$. Assume that $\lim_{x\to x_0^+}f^\prime(x)=\infty$. Does it imply that $f^{'}_{+}(x_0)$ does not exist?

I know that in general the derivative need not behave well. I have as an example the derivative of $x^2\sin(\frac{1}{x})$ which does not have a limit at $x_0=0$, yet the function is differentiable at $x_0=0$ (if forced to be continuous at $x_0$). This example, though, has a bounded derivative. I was wondering if relieving this requirement forces $f$ to not be differentiable.

$\endgroup$
  • $\begingroup$ doesn't your hypotheses imply that also $f'_+(x_0)$ is $\infty$? $\endgroup$ – dfnu Feb 14 '19 at 19:13
0
$\begingroup$

Hint: Let $x > x_0$. From MVT $f(x)-f(x_0)=(x-x_0)f’(c_x)$ where $x_0 < c_x < x$.

Thus, as $x$ goes to $x_0$, $f’(c_x)$ goes to $+\infty$...

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.