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While solving a problem, I found the statement below

If $f:[0,+\infty)\rightarrow \mathbb{R}$ and $f$ is differentiable on $(0,+\infty)$, then $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$ if and only if $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=+\infty$

I don't know whether this statement is true or not, but I found a solution for it. Here is my solution:

$\bullet$ If $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$ then there exists an $X$ such that for all $x>X:f'(x)>1$. By Lagrange theorem, for every $x>X$ there is $C_x\in(X,x)$ such that $$f(x)-f(X)=(x-X)f'(C_x)>x-X$$ Thus $f(x)$ tends to infinity when $x$ tends to infinity. By L'Hopital Rule: $$\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$$ $\bullet$ If $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=+\infty$ then it's easy to prove $\displaystyle \lim_{x\rightarrow +\infty}f(x)=+\infty$. By L'Hopital Rule, we get $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$

From the above proof, we can also conclude that if $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$ and by Lagrange theorem, for every $x>0$ there exists $c_x\in (0,x)$ such that $$f(x)-f(0)=xf'(c_x)$$ then the set of all $c_x$ (with all $x>0$) does not have a upper bound.

Is there any wrong with my statement and proof? Thank you so much.

Tien Kha Pham
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  • In your first part, there is a problem: dividing $f(x)-f(X) > x-X$ by $x$ only leads you to $f(x) > 1 + o(1)$, which is not enough to conclude that $f \to \infty$. You could maybe try to replace "$1$" by an arbitrary $K > 0$. – Alexandre Halm Jan 06 '15 at 06:24
  • And in your second part, $f(x) \to \infty \implies f'(x) \to \infty$ is obviously wrong, as the counter-example $f(x)=x$ shows. – Alexandre Halm Jan 06 '15 at 06:25
  • @AlexH. Thank you for answering my question. Maybe you have some misread between $f(x)$ and $f(x)/x$. In my first part, I only get $\lim f=+\infty$ from $f(x)-f(X)>x-X$, then I use L'Hopital Rule for $f(x)/x$. – Tien Kha Pham Jan 06 '15 at 09:17

1 Answers1

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You are applying l'Hôpital's rule incorrectly. Rather, the conclusion should be that if $f'(x)\to\infty$, then indeed $\displaystyle\lim_{x\to\infty}\frac{f(x)}x=+\infty$. We are using here the form of l'Hôpital's rule (as presented, for instance, in Rudin's book, see Theorem 5.13) that states that if $a$ is in the extended reals, $\lim_{x\to a}g'(x)/h'(x)=L$ exists (in the extended reals), and $\lim_{x\to a}h(x)=+\infty$, then also $\lim_{x\to a}g(x)/h(x)=L$. Note that there is no need to assume that $\lim_{x\to a}g(x)=\infty$ as well.

On the other hand, if $\displaystyle \lim_{x\to\infty}\frac{f(x)}{x}=\infty$, we cannot conclude that $\lim_{x\to\infty}f'(x)=\infty$. By l'Hôpital's rule, if $\lim_{x\to\infty}f'(x)$ exists, then it must be $\infty$. But the limit may fail to exist. For instance, we could have $f$ growing exponentially except that it is constant in little intervals around each integer. For any such $f$ we have $\lim_{x\to\infty}f(x)/x=+\infty$; however $\limsup_{x\to\infty}f'(x)=+\infty$ but $\liminf_{x\to\infty}f'(x)\le0$, so the limit does not exist.

Andrés E. Caicedo
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