I'm trying to solve this question:
Let $f:[0,+\infty) \to \mathbb{R}$ be derivable and $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x\to \infty}\frac {f(x)}{x}=L$.
I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if $\lim_{x\to +\infty} f(x)=+\infty$.
I need help.
Thanks a lot.
As mentioned in the comments, L'Hôpital's rule is applicable, and readily yields the result.
Here's a proof that does not appeal to L'Hôpital's rule (though it essentially uses the same proof as that result).
First assume $L$ is finite. Let $\epsilon>0$.
By the Mean Value Theorem and the hypothesis that $\lim\limits_{x\rightarrow\infty} f'(x)=L$, there is a number $a$ so that for all $x>a$ $$\tag{1} L-\epsilon < {f(x)-f(a)\over x-a} <L+\epsilon. $$ Since ${x-a\over x}=1-{a\over x}>0 $ for $x>a$ it follows from $(1)$ that $$\tag{2} (L-\epsilon)\Bigl(1-{a\over x}\Bigr) < {f(x)\over x}- {f(a)\over x} < (L+\epsilon)\Bigl(1-{a\over x}\Bigr), $$ for $x>a$.
Since $\lim\limits_{x\rightarrow\infty} {f(a)\over x}=\lim\limits_{x\rightarrow\infty }{a\over x}=0$, it follows from $(2)$ that $$ \textstyle \limsup\limits_{x\rightarrow\infty} {f(x)\over x}\le L+\epsilon \ \ \text{and}\ \ \liminf\limits_{x\rightarrow\infty} {f(x)\over x}\ge L-\epsilon. $$
Since $\epsilon$ was arbitrary, we have $$\textstyle \limsup\limits_{x\rightarrow\infty} {f(x)\over x}\le L \ \ \text{and}\ \ \liminf\limits_{x\rightarrow\infty} {f(x)\over x}\ge L; $$ whence the result follows.
If $L=\infty$, the argument is similar; but one starts with an arbitrary $M>0$, and then finds an $a$ so that for all $x>a$, $$\tag{1} {f(x)-f(a)\over x-a} >M. $$ Then $$\ {f(x)\over x}- {f(a)\over x} >M\Bigl(1-{a\over x}\Bigr). $$ From this it follows that for $x$ sufficiently large ${f(x)\over x}\ge M\cdot(1/2)-1/2. $ Since $M$ was arbitrary, we have $\lim\limits_{x\rightarrow\infty}{f(x)\over x}=\infty$.
The argument for $L=-\infty$ proceeds as in the argument for $L=\infty$.
