If $\lim_{x \to \infty} f(x) - xf'(x)$ exists, does $\lim_{x \to\infty} f'(x)$ exist as well?

Solution 1:

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.

Solution 2:

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?