# If $\lim_{x\to x_0^+}f^\prime(x)=\infty$ then $f^{'}_{+}(x_0)$ does not exist.

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.

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

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 $$+\infty$$...