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Here is the exercise that I can not find the proof at present: Let $f:[0,\infty)\to\mathbb{R}$ be a differentiable function satisfying $\lim_{x\to+\infty}\frac{f(x)}{x^2}=1, \text{ and } \lim_{x\to+\infty}f'(x)=L$. Show that $L=+\infty$. This exercise is extracted from Giaqinta & Modica's TextBook: Mathematical Analysis, Foundations of one variable, Page143, 3.86.

I have tried like the following: If $L\in\mathbb{R}$, then there exist $M>0, X>0,$ such that $|f'(x)|\leq M, \forall x\geq X.$ Thus for each $x>X$, $\exists \eta\in (X,x)$, such that $f(x)-f(X)=f'(\eta)(x-X), $ and then $$\frac{f(x)}{x^2}=\frac{f(X)}{x^2}+(\frac{1}{x}-\frac{X}{x^2})f'(\eta)\to 0, \text{ as } x\to+\infty,$$ which contradicts the condition $\frac{f(x)}{x^2}\to 1, \text{as} x\to+\infty.$ Therefore $L\in\{+\infty, -\infty\}.$ But How to conclude that $L\not=-\infty$?

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Surely the first condition on $f$ implies $f'\gt0$ a lot, making $f'\to-\infty$ impossible. – Gerry Myerson Sep 7 '12 at 3:11
This might not directly answer your question, but you should note that $f(x) \to \infty$. By L'Hopital's rule you know that the limit of $f'(x)$ as $x \to \infty$ is the same as the limit of $2x$ as $x\to \infty$. – Shankara Pailoor Sep 7 '12 at 3:14
Actually I think l'Hopital's Rule can not be applied here if you could not confirm that $\lim_{x\to+\infty}\frac{f'(x)}{2x}$ exists. – nuage Sep 7 '12 at 3:29
up vote 1 down vote accepted

If $L = -\infty$, then for sufficiently large $x$, $f'(x) < -1$, and so $f(x) < k - x$ where $k$ is some constant. This implies $\frac{f(x)}{x^2} < \frac{k - x}{x^2}$ for large $x$. Therefore, $\limsup_{x\to\infty} \frac{f(x)}{x^2} \le 0$.

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Tunococ's proof of $L\not=-\infty$ is very wonderful and neat. Thank you very much! – nuage Sep 7 '12 at 4:43

Added Much simpler argument:

It is clear that $\lim_{x \to \infty} f(x)= \infty$.

Since $\lim_{x \to \infty} f'(x)=L$, exists, and e are in teh $\frac{\infty}{\infty}$ case, by L'Hospital we get that

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


$$L=\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \left( \frac{f(x)}{x^2} \right)x = 1 \cdot \infty=\infty$$

You can use the Mean Value Theorem:

For each $n$ there exists a $c_n$ so that

$$\frac{f(2n) -f(n)}{n}= f'(c_n)$$

Now, with $\epsilon=\frac{1}{5}$, there exists an $N$ so that for all $n >N$, we have

$$f(2n)>4n^2(1-\frac{1}{5}) \,;\, f(n)< n^2 (1+\frac{1}{5})$$

Thus, for all $n >N$we have

$$f'(c_n) >2n (*)$$

Letting $n \to \infty$, since $c_n >n$ we have $c_n \to \infty$, and hence $f'(c_n) \to L$. Then, (*) implies that $L=\infty$.

P.S. This argument can probably be simplified.

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I am not so great with mathematical formalism, but keep in mind the asymptotic behavior of the quotient $f(x)/x^2$ means that the polynomial representation of $f(x)$ has a degree of 2 and a leading coefficient 1. (This is a property of rational functions).

That is $f(x) = x^2 + ax + b + ...$

As a result $f'(x) = 2x + a + ...$, and its limit as $x \to +\infty$ is therefore $+\infty$

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