Let $f(x)$ be a differentiable function in $[a,b]$. Suppose $\lim _{x\to b^-}f\left(x\right)=\infty $ and $f^\ {'}(x)$ is monotonic, prove $\lim _{x\to b^-}f^\ {'}\left(x\right)=\infty $. Is the conclusion still true when $b=\infty?$

My try:

Let $x\in(a,b)$, so by MVT we get there exists $c\in(a,x)$ such that


The derivative must be monotonic increasing, otherwise we will get a contradicition to $\lim _{x\to b^-}f\left(x\right)=\infty $.

So we have $f^{\ '}(x)>f^{\ '}(c)$, now taking limits on both sides: $$\lim _{x\to b^-}f^{\ '}(x)\ge \lim _{x\to b^-}\frac{f\left(x\right)-f\left(a\right)}{x-a}=\infty $$And we get the result. I think it is not true anymore when $b=\infty$, but not quite sure how to prove it.

Is my proof correct?

  • 2
    $\begingroup$ If $f$ is differentiable on $[a,b]$ then it is continuous on this compact interval, and so it is bounded on $[a,b]$. Thus it would not be possible to have $\lim_{x \to b^-} f(x) =\infty$. Did you mean $f$ is differentiable on $(a,b)$? $\endgroup$ – joeb Feb 23 '17 at 16:09
  • $\begingroup$ @joeb It says $[a,b]$ in the textbook, I guess it's a typo? $\endgroup$ – Itay4 Feb 23 '17 at 16:13
  • $\begingroup$ Probably a typo. They might mean $f$ is differentiable on $[a,b)$. The phrase "what if $b=\infty$" seems to support this $\endgroup$ – joeb Feb 23 '17 at 16:17

The proof is correct if you mean differentiable in $[a,b)$ (obviously $f$ can't be continuous in $b$ if it diverges to $\infty$ there), as is the hunch for $b=\infty$ (not terribly formal, but we get the idea). Just consider $f(x)=\sqrt{x}$ in $[1,\infty)$ for example.


I think you need to change the question a bit.

Let $f$ be differentiable in $(a, b)$ and $\lim_{x \to b^{-}}f(x) = \infty$. If $f'$ is monotone on $(a, b)$ show that $\lim_{x \to b^{-}}f'(x) = \infty$. Does the same conclusion hold if $b = \infty$?

Since $f'$ is monotone, either $f'$ tends to a limit as $x \to b^{-}$ or tends to $\pm\infty$. If it tends to $-\infty$ then we can see that as $x \to b^{-}$ the derivative $f'$ is negative and hence $f$ is decreasing and hence there is no way it can tend to $\infty$. Thus we have either $f'(x) \to L$ or $f'(x) \to \infty$. Let us then assume that $f'(x) \to L$ as $x \to b^{-}$. Then we have number $c \in (a, b)$ such that $f'(x)$ is bounded in $(c, b)$. And then by mean value theorem we can see that $$f(x) = f(c) + (x - c)f'(d)$$ for $x \in (c, b)$ and some $d \in (c, x)$. Letting $x \to b^{-}$ we get an obvious contradiction as LHS tends to $\infty$ and the RHS is bounded. Thus the only option left is that $f'(x) \to \infty$.

When $b = \infty$ then you have many examples to convince that $f(x) \to \infty$ does not imply $f'(x) \to \infty$ even if $f'$ is monotone. One such example is $f(x) = \log x$.

  • $\begingroup$ Did you mean that $f^{'}$ is negative if it tends to $-\infty$? $\endgroup$ – Itay4 Feb 23 '17 at 16:29
  • $\begingroup$ @Itay4: Thanks for catching typo. $\endgroup$ – Paramanand Singh Feb 23 '17 at 16:30
  • $\begingroup$ @Itay4: I did not understand why you wrote that $f'$ must be monotonically increasing in your solution? $\endgroup$ – Paramanand Singh Feb 23 '17 at 16:31
  • $\begingroup$ I did not show the proof here, but I contradicted the fact it is monotonically decreasing the same way you did. $\endgroup$ – Itay4 Feb 23 '17 at 16:34
  • $\begingroup$ @Itay4: ok then your solution is correct. I added my answer only to explicitly show that $f'$ increases to $\infty$ and thought that your solution assumed this without any proof. $\endgroup$ – Paramanand Singh Feb 23 '17 at 16:36

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.