If f: [0, +infinity) --> R is differentiable and lim f'(x) = L (as x tends to +infinity) then lim [f(x)/x] = L (as x tends to +infinity).
help, anyone?
thanks
Use L'Hopital rule.
Fernando Revilla
Is it forbidden?
Fernando Revilla
Well, in that case a possibility is to translate the proof of the L'Hopital rule with $\displaystyle g(x)=x$ . Nobody will dare to say you have used it. Take into account that:
$\displaystyle \dfrac{f(x)-f(0)}{x-0}=f'(\xi),\quad \xi \in (0,x)$
by the Lagrange's theorem
Fernando Revilla
Use:
$\displaystyle \dfrac{f(x)}{x}-\dfrac{f(0)}{x}=f(\xi)$
and
$\displaystyle \displaystyle\lim_{\xi \to +\infty}f(\xi)=L$
Fernando Revilla
P.S. Reading quickly at the OP, I "instinctively" saw $\displaystyle \lim_{x \to +\infty}f(x)=+\infty$ .
Please, correct if I'm wrong.
What you're saying is that by the Mean Value Theorem we can find $\displaystyle \xi \in (0,x)$ such that $\displaystyle \dfrac{f(x)}{x}-\dfrac{f(0)}{x}=f'(\xi)\quad$, for all $\displaystyle x>0$. Yet I still don't see how we can infer that $\displaystyle \displaystyle\lim_{x \to +\infty}\dfrac{f(x)}{x}=L$ since $\displaystyle x \to +\infty$ does not imply $\displaystyle \xi \to +\infty$. Thus we cannot use the fact that $\displaystyle \displaystyle\lim_{\xi \to +\infty}f'(\xi)=L$.
If the denominator goes to infinity at the limit point, there's no need for the numerator to go to infinity as well, even though in many textbooks it is written that it is necessary.
Check the proof you have of L'hospital's theorem for the case $\displaystyle \frac{\infty}{\infty}$. Does it really use the fact that the numerator goes to infinity?
As FernandoRevilla pointed, you can rewrite the proof of L'hospital's rule to this particular case if you are not allowed to directly use the theorem.