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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

$$f\:'\left(c\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}$$

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?

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    $\begingroup$ 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
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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.

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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 $f(x) = \log x$.

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  • $\begingroup$ 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

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