If $\lim_{x \to \infty} \frac{f'(x)}{x}=2$ does it follow that $\lim_{x \to \infty} \frac{f(x)}{x^2}=1$?

I need to show that the following statement is true or false. $$\displaystyle\lim_{x \to \infty} \frac{f'(x)}{x}=2 \Rightarrow \displaystyle\lim_{x \to \infty} \frac{f(x)}{x^2}=1$$

I considered $f(x)=x^2$, and it seems that the above statement is true. If so, how do I go about proving this? Or maybe there is a counterexample?

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Do you know l'Hôpital's theorem? – egreg May 27 '13 at 22:31
@egreg Yes, but I have only used it to in the form $\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=\lim_{x\to\infty}\dfrac{f'(x)}{g'(x)}$. I didn't realize it could be the other way round. – mathmo May 27 '13 at 22:45
It is the other way around! When using l'Hôpital's theorem, you need to know that the limit of the ratio of the derivatives exists. – egreg May 27 '13 at 22:50
Oh my. Thanks for pointing that out. – mathmo May 27 '13 at 22:55

$\lim\limits_{x\to\infty}\dfrac{f'(x)}{x}=2$, therefore $\lim\limits_{x\to\infty}{f'(x)}$ equals $\infty$. Now apply De L'Hospital's rule to the second limits and the result follows.

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L'Hôpital's rule can be proven true if either both numerator and denominator go to $0$, or if only the denominator goes to $\infty$. You can see a proof of this slightly general assertion in Rudin's Principles. – Pedro Tamaroff May 27 '13 at 22:37
That's a nice generalization. I thought the only cases were $\pm 0, \pm\infty$ for both numerator and denominator. – Dimitrios Nt May 27 '13 at 22:40
Thank you. I didn't know it could be the other way round. Usually I encounter the function first, then the derivative, $\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=\lim_{x\to\infty}\dfrac{f'(x)}{g'(x)}$. – mathmo May 27 '13 at 22:47
But you need to argue—or at least observe!—that $f(x)\to\infty$ before you can use L'Hôpital "backwards." – Ted Shifrin May 28 '13 at 1:51

Since

$$\lim_{x \to \infty} \frac{f'(x)}{x} =2,$$

we have that $f'(x) > x$ for sufficiently large $x$. With this, we may show that $f(x)$ increases without bound (i.e. $\lim_{x \to \infty} f(x) = \infty$). Thus the limit of $f(x)/x^2$ meets the hypotheses of l'Hopital, so that $$\lim_{x \to \infty} \frac{f(x)}{x^2}= \lim_{x \to \infty} \frac{f'(x)}{2x} = \frac{1}{2} \lim_{x \to \infty} \frac{f'(x)}{x} = 2 \cdot \frac{1}{2}=1.$$

As Peter Tamaroff mentions in the comments elsewhere on this page, the assumption (in l'Hopital's rule) that our numerator tends to infinity is actually superfluous: it suffices for the denominator to go to infinity.

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