Let $f:\Bbb R \to \Bbb R$ be a differentiable function. If $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = + \infty $, it is always true that $\mathop {\lim }\limits_{x \to + \infty } f'(x) = + \infty $? How about the converse?

For example, $\mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{x} = 0$ is finite, then we can see $\mathop {\lim }\limits_{x \to + \infty } (\ln x)' = 0$ is finite. $\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}}}{x} = + \infty $ so $\mathop {\lim }\limits_{x \to + \infty } ({x^2})' = \mathop {\lim }\limits_{x \to + \infty } x = + \infty $. So the claim seems good to me, but I don't know how to actually prove it. $\mathop {\lim }\limits_{x \to + \infty } f'(x) = \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$, I don't know how to deal with this mixed limit. Also since the limits in the proposition diverges, it looks like mean value theorem sort of thing cannot apply here.

  • $\begingroup$ Can someone explain expalain to me why we have to completely opposed answer on this post ? and both have highest upvotes $\endgroup$ – Guy Fsone Jan 29 '18 at 16:28
  • $\begingroup$ @GuyFsone Could you explain how they're opposed? $\endgroup$ – MathematicsStudent1122 Jan 30 '18 at 19:03
  • $\begingroup$ The first answer prove a theorem but the second has a counterexample $\endgroup$ – Guy Fsone Jan 30 '18 at 19:43
  • $\begingroup$ @GuyFsone The second answer isn't a counterexample to the theorem I have in the yellow box. $\lim \sup$ is different from $\lim$. I actually make explicitly clear in my answer that $$\frac{f(x)}{x} \to \infty$$ doesn't imply $f' \to \infty$. $\endgroup$ – MathematicsStudent1122 Jan 31 '18 at 19:58
  • $\begingroup$ @MathematicsStudent1122 ok thanks you are right I did not pay enough attention $\endgroup$ – Guy Fsone Jan 31 '18 at 20:00

The "left to right" of the biconditional is true. As noted in another answer, we can use L'hopital. But I will utilize a direct approach. We need to show that for arbitrarily large $M$, we have for sufficiently large $x$ the inequality $\frac{f(x)}{x} > M$.

By assumption, for any arbitrarily large $M$ we have $f'(x) > 2M$ when $x>x_0$. This means $f(x) \geq f(x_0) + 2M(x-x_0)$ for $x > x_{0}$. Note also that there is an $x_1$ such that for all $x > x_1$, $f(x_0) + 2M(x-x_0) > {Mx}$. Hence, we can see that for $x > \max\{x_1, x_0\}$ we have $$\frac{f(x)}{x} \geq \frac{f(x_0) + 2M(x-x_0)}{x} > \frac{Mx}{x} = M $$

The "right to left" of the biconditional is false. Consider $f(x) = x^2(\sin x + 2)$. This is positive and bounded below by $x^2$, hence $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$ but $f'$ oscillates as $x \to +\infty$.

We can say something weaker, however, namely the following

Theorem: Let $f \in C^1(\mathbb{R})$ such that $$\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$$ Then we have $$\limsup_{x \to +\infty} \ f'(x) = +\infty$$

To prove this, first note that for $f$, we can assume $f(0) = 0$ without any loss of generality. Indeed, define $g(x) = f(x) - f(0)$ and note $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty \Longleftrightarrow \lim_{x \to +\infty} \frac{g(x)}{x} = +\infty$ and also $f' = g'$.

We can prove by contradiction. Suppose the $\lim \sup$ is finite or $-\infty$. This means $f'$ is bounded above in $[M, +\infty)$ for some $M>0$. Since $f'$ is continuous, by the extreme value theorem it is bounded above in $[0,M]$, and hence it is bounded above in $[0, +\infty)$. By the mean value theorem, we have that $\frac{f(x)}{x} = f'(\alpha)$ for some $\alpha$ in $[0, x]$. Letting $x \to +\infty$ we can see that $f'(\alpha)$ takes on arbitrarily large positive values, which contradicts the fact that $f'$ is bounded above in $[0, +\infty)$.

This can probably be modified so that the $C^1$ condition can be relaxed (e.g., to allow for cases where $f'$ is discontinuous), but I'm not sure how to do that.

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  • $\begingroup$ Thanks a lot! This is very helpful. Will the converse the theorem hold? That is $\limsup_{x \to +\infty} \ f'(x) = +\infty$ implies $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$? Thanks! $\endgroup$ – Tuyet Mar 14 '17 at 4:13
  • $\begingroup$ @Tuyet No, $f(x) = x\sin x$. $\endgroup$ – MathematicsStudent1122 Mar 14 '17 at 4:15
  • $\begingroup$ I am so dumb :( I cannot figure out why $f'(x) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f({x_0})}}{{x - {x_0}}} > 2M$ for $x>x_0$ implies $\frac{{f(x) - f({x_0})}}{{x - {x_0}}} > 2M$ for $x>x_0$. I can see here for every $x_1$ that is sufficiently near each $x > x_0$, then we have $\frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}}>2M$ by limit's property. $\endgroup$ – Tuyet Mar 14 '17 at 4:47
  • $\begingroup$ how can we say from here that $x>x_0$ implies $\frac{{f(x) - f({x_0})}}{{x - {x_0}}} > 2M$? $\endgroup$ – Tuyet Mar 14 '17 at 4:53
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    $\begingroup$ You can indeed get rid of the $C^1$ assumption. If $\lim_{x\to +\infty}\frac{f(x)}{x}=+\infty$ then $\lim_{x\to +\infty}f(x)=+\infty$. So for your $M$, $\lim_{x\to +\infty} \frac{f(x)}{f(x)-f(M)}=1$ and $\lim_{x\to +\infty} \frac{x}{x-M}=1$. This means that your could use the mean value theorem on $[M, x]$ to get that $f'(\alpha)=\frac{f(x)-f(M)}{x-M}\to +\infty$. $\endgroup$ – Peradventure Mar 14 '17 at 6:06
  1. It is true that if $\lim_{x\to +\infty} f'(x)=+\infty$ then $\lim_{x\to +\infty} \frac{f(x)}{x} = + \infty$. This can be proved by using the methods given by Dr.MV's answer to your question.
  2. It is in general false that if $\lim_{x\to +\infty} \frac{f(x)}{x}=+\infty$ then $\lim_{x\to +\infty} f'(x)=+\infty$.

Counter-example: Let $f(x)=x^2 + \sin x^3$. Then $\lim_{x\to +\infty} \frac{f(x)}{x} =+\infty$. But $f'(x) = 2x+3x^2\cos x^3$. Note that $f'(x)$ is continuous for all $x\in \mathbf{R}$, but since the sign of $\cos x^3$ could vary as $x$ goes to $+\infty$, $\lim_{x\to +\infty}f'(x)$ doesn't exist and is not $+\infty$.

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If $\lim_{x\to \infty}f'(x)$ exists, then from L'Hospital's Rule we have

$$\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to \infty}f'(x)$$

regardless of whether $\lim_{x\to \infty}f(x)$ exists or not (See the note that follows Case 2 of THIS ARTICLE).

Hence, if $\lim_{x\to \infty}f'(x)=\infty$, then $\lim_{x\to \infty}\frac{f(x)}{x}=\infty$ also.

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  • $\begingroup$ Thanks so much for your help! For mean value theorem proof, how do we guarantee $\xi \to \infty$ as $x \to \infty$? Many thanks! $\endgroup$ – Tuyet Mar 14 '17 at 3:44

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