$f(x)$ is positive, continuous, monotone and integrable in (0,1]. Is $\lim_{x \rightarrow 0} xf(x) = 0$? I'm having trouble with this question from an example test.
We have a positive function $f(x)$ that's monotone, continuous and integrable in $(0,1]$. Is $\lim_{x \rightarrow 0} xf(x) = 0$?
Progress
The only problematic case seems to be when $f(x)$ is unbounded and monotonic decreasing. For that case, I found out that $xf(x)=\int_{0}^{x} f(x)dt$ and that $0\leq xf(x)\leq \int_{0}^{x} f(t)dt$. From here I'm not sure how to go on.
Thanks!

If $f$ is increasing, then $\lim_{x \to 0^+} f(x) = C < +\infty$ exists, so that the result is trivial. 
If $f$ is decreasing, $c < x$ implies $f(c) \ge f(x)$. Now by the mean value theorem for integrals, for every $0 < x < 1$, there exists $0 < c(x) < x$ such that 
$$
\int_0^x f(t) \, dt = f(c(x))x. 
$$
This gives
$$
0 \le x f(x) \le x f(c(x)) = \int_0^x f(t) \, dt \to 0.
$$
Hope that helps,
