Prove that $\frac{x}{f(x)}$ is decreasing Suppose f>0 on $(1,\infty)$ and $\lim_{x\to\infty}{\frac{xf'(x)}{f(x)}} \in (1,\infty)$. Prove that $\frac{x}{f(x)}$ is decreasing on $(b,\infty)$ for some b>1.
I try to build a contradiction by showing it is not decreasing, then ...
But I don't have some concrete idea so far. Anyone can give me some hint?
 A: Hint: Note that $\left(\tfrac{x}{f(x)}\right)'=\tfrac{f(x)-xf'(x)}{f(x)^2}$. Since the denominator is positive by hypothesis, this derivative has the same sign as the numerator $f(x)-xf'(x)$.
A: I assume $f$ is differentiable.
Note that, for $g$ defined by $g(x) = \frac{x}{f(x)}$ on $(1,\infty)$ (well-defined as $f>0$ by assumption), we have that $g$ is differentiable, with
$$
g'(x) = \frac{f(x)-x f'(x)}{f(x)^2} = \frac{1}{f(x)}\left(1-\frac{xf'(x)}{f(x)}\right)
$$
for every $x>1$. By our other assumption, there exists $\ell>1$ such that
$$
\lim_{x\to\infty} \left(1-\frac{xf'(x)}{f(x)}\right) = 1-\ell < 0
$$
and so there exists some $b>1$ such that $\left(1-\frac{xf'(x)}{f(x)}\right) \leq \frac{1-\ell}{2} < 0$ for all $x>b$.
It follows that $g'(x) < 0$ for every $x>b$, allowing you to conclude.
A: Since you are given $\displaystyle \lim_{x\to+\infty} \frac{x f'(x)}{f(x)} > 1$ then I assume $f$ must be differentiable, since otherwise that given condition just wouldn't make sense.
Consider then $$\frac d{dx}\left(\frac x{f(x)}\right) = \frac{f(x) - xf'(x)}{f(x)^2} = \frac1{f(x)}\left(1 - \frac{xf'(x)}{f(x)}\right).$$
Since $\displaystyle \lim_{x\to+\infty} \frac{x f'(x)}{f(x)} > 1$, there must be some value of $b$ such that...
Can you take it from here?
