If $\lim_{x \to \infty} f(x) - xf'(x)$ exists, does $\lim_{x \to\infty} f'(x)$ exist as well? Let $f(x)$ be a differentiable function on $(0, \infty)$ with $\lim_{x\to \infty} f(x) - xf'(x) = L\in \mathbb{R}$. I'm trying to prove or disprove that $\lim_{x\to\infty} f'(x)$ exists as well. Here's what I have so far: if the limit does exist, it must also equal $\lim_{x\to\infty} \frac{f(x)}{x}$ (divide the limit condition by $x$). Then I rewrote the condition as $x^2 \frac{d}{dx} \frac{f(x)}{x} \to L$, and from this I expect (using some $1/x^2$ asymptotic argument) my statement to be true - but I'm not sure how to rigorously proceed with this argument.. Can someone provide some next steps, or a counter example?
 A: Yes, $\lim_{x \to\infty} f'(x)$ exists under those conditions. In fact it suffices to require that the function
$$
 r(x) = f(x) - xf'(x)
$$
is bounded for $x \to \infty$, i.e. that
$$
 M = \sup \{ |r(x)| : x \ge 1 \} < \infty \, .
$$
Proof:
$$
 f'(x) = \frac{f(x)}{x} - \frac{r(x)}{x}
$$
so that the claim is equivalent to the existence of $\lim_{x \to\infty} \frac{f(x)}{x}$. Here we can apply the Cauchy criterion to
$$
 F(x) = \frac{f(x)}{x} \, .
$$
For $x \ge 1$ is
$$
 |F'(x)| = \left|\frac{xf'(x) - f(x)}{x^2}\right| = \frac{|r(x)|}{x^2} \le \frac{M}{x^2} \, .
$$
It follows that for $1 \le x < y$
$$ 
 |F(x) - F(y)|  \le \int_x^y \frac{M}{t^2} \, dt = M \left( \frac 1x - \frac 1y \right) < \frac Mx \, .
$$
(This is a straight-forward estimate if the fundamental theorem of calculus can be applied to $F$, e.g. if $F'$ is continuous or Riemann integrable. For the general case see $|f'(x)| \le g(x)$ implies $|f(b) - f(a)| \le \int_a^b g(x) dx$, without assuming $f'$ to be integrable.).
So for every $\epsilon > 0$
$$
 x, y > \frac{M}{\epsilon} \implies  |F(x) - F(y) | < \epsilon 
$$
and that completes the proof.
A: Here is a path to a proof. Denote $f(x) = x\cdot g(x)$ Then equivalently what we are asking is that if the limit
$$\lim_{x\to\infty}-x^2g'(x) = L$$
exists, then does the limit
$$\lim_{x\to\infty}g(x)+xg'(x)$$
exist? Further we have by squeeze theorem that
$$\lim_{x\to\infty} xg'(x) = 0$$
so we can simplify this to the question does the limit
$$\lim_{x\to\infty}g(x)$$
exist given the condition above?
