# Proving the existence of the Limit of a function as x goes to infinity.

Suppose that $f:[0, \infty) \to \Re$ is a $C^2$-function, i.e., a twice continuously differentiable function. There exists a linear function $ax + b$ with $a > 0$ such that

$f(x) \leq ax + b$ for all $x \geq 0$.

Also its derivative $f'(x)$ is strictly positive and bounded, i.e., there exists a $M > 0$ for which $f'(x) < M$ for all $x \geq 0$. Furthermore,

$$f'(x) = \frac{c}{x g(x)},$$ where $c>0$ is a constant and $g(x)$ is a function with the property $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} g'(x) = 0$.

Hence $\lim_{x \to \infty} x \; f'(x)$ diverges to $\infty$.

I'm trying to prove (or disprove) the following statement:

"$\lim_{x \to \infty} f'(x)$ exists and the limit value is $a$."

I believe that this conjecture is true but don't know how to prove it. Any help is greatly appreciated!

• Do you mean $\mathbb{R}$? – Eoin Nov 28 '14 at 21:39
• If $\displaystyle\lim_{x\to\infty}g(x) = 0$, then $\displaystyle\lim_{x\to\infty}xg(x) = 0$, and so $\displaystyle\lim_{x\to\infty}f'(x) = \lim_{x\to\infty}\dfrac{c}{xg(x)}$ can't have a finite value $a$. EDIT: Ignore this comment, it is incorrect. – JimmyK4542 Nov 28 '14 at 21:45
• @JimmyK4542 the first implication is false, take $g(x) = \frac{1}{\sqrt x}$. – TZakrevskiy Nov 28 '14 at 21:48
• ^Thanks for correcting that. I was thinking of $x \to 0$ for some reason. – JimmyK4542 Nov 28 '14 at 21:49
• Take $f=kx$ with $k \le a$, then with $c=1$ you obtain $g(x)=\frac{1}{kx}$ (i.e. satisfies your hypothesis), yet $\lim f'(x)=k$. Hence, even you have the existence of the limit, you can not guarantee that $\lim f'(x)=a$. – TZakrevskiy Nov 28 '14 at 21:51

Take $$f(x) = 2x+\sin x\le 3x+1.$$ This function is strictly increasing and its derivative is bounded.
Yet $$f'(x) = 2+\cos x = \frac{1}{x\cdot\frac{1}{x(2+\cos x)}},$$i.e. $$g(x)=\frac{1}{x(2+\cos x)}.$$ Obviously, $$\lim g(x)=\lim g'(x)=0,$$ so all your conditions are satisifed.
However, $\lim f'(x)$ does not exist.