$\lim_{x\to a+}f(x)=-\infty$ therefore $\lim_{x\to a+}f'(x)=\infty$?

Solution 1:

Suppose $f$ is chosen so that it is differentiable on $(a,b)$ and $f'$ is uniformly continuous (that is to say for all $\epsilon>0$ there exists an $\delta>0$ so that if $|x-y|<\delta$ then $|f'(x)-f'(y)|<\epsilon$ then this holds true. It is also true if $f'$ is monotonic.

In fact it is easy to see that if as long as $f'$ exists and it is bounded from above (so $f'\leq C$) then $\liminf_{x\to a}f(x)$ is bounded from below: We get for any $x,\epsilon$ that for $|h|$ sufficiently small we have $$ \frac{f(x)-f(x-h)}{h} < C+\epsilon $$ This implies $f(x-h) > f(x)-(C+\epsilon)h$.

Now fix some $r\in(a,b)$ and take $l$ infimal so that $f(x)\geq f(r)-(C+\epsilon)(r-x)$ for all $x\in(l,r)$. Then by limiting we get $$ f(l) \geq f(r) - (C+\epsilon)(r-l) $$ Suppose $l\neq a$. Then we can find an $\rho>0$ so that for $|h|<\rho$ we get $$ f(l-h) > f(l)-(C+\epsilon)h > f(r) - (C+\epsilon)(h+r-l) = f(r) - (C+\epsilon)(r-(l-h))$$ which contradicts the infimality of $l$.

But thus $f$ is bounded towards $a$ from below by $f(r)-(C+\epsilon)(r-a)$.

So by contraposition if $f$ limits to $-\infty$ it is not bounded from below and thus $f'$ cannot be bounded from above. But this only means that there exists one sequence $x_n\to a$ so that $f'(x_n)\to\infty$.

If $f'$ is uniformly continuous this is sufficient to say that any sequence $a_n\to a$ satisfies this (as if $|x_n-a|<\delta$ and $|y_n-a|<\delta$ (which is true for $n$ sufficiently large) we get $|x_n-y_n|<\delta$, and thus $f'(y_n)>f'(x_n)-\epsilon$). As you can see we do not in fact need uniformly continuous here, it it sufficient if at least for one $\epsilon>0$ there is such a $\delta>0$.

Similarly this holds true for $f'$ monotonic close to $a$.

If this condition does not hold it might be that $f'$ does not in fact take a limit (in some sense). For example:

https://sagecell.sagemath.org/?z=eJxFjUEKwjAQAO-F_iG3bmy0iVBv8SMiEkyiC20NSYT19268eJ0ZGB-BpIWCG5iJ5GjkRH2XllcFH5U-GGVk37lS3muA44725qwZRGEFbjU8squhlcTtrIhVyixgwRUrMLdaecx2GAfJkvv0xHu5uZzdBy6_Ufto3U5K_M_6xOAqv3hPLgM=&lang=sage&interacts=eJyLjgUAARUAuQ==