Limit of $x\log x\cdot f(x)$ when $f$ is integrable function Suppose that $f\in L^1(0,+\infty)$ is a monotone decreasing, positive function. Prove that $$\lim_{x \to +\infty}x(\log x)\cdot f(x)=0.$$
The assertion is not true. Consider 
$$
f_0=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\mathbf 1_{[0,\mathrm e^{n^2}]},
$$
then $f_0$ is nonincreasing, $g_0(x)=f_0(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g_0(\mathrm e^{n^2})\geqslant1$ for every $n\geqslant1$, and the integral of $f_0$ is $\sum\limits_{n\geqslant1}n^{-2}$, which is finite, hence $f_0$ is integrable. 

One can modify this example slightly to get a decreasing function $f$, for example 
$$
f(x)=\sum\limits_{n\geqslant1}(n^2\mathrm e^{n^2})^{-1}\exp(-x\mathrm e^{-n^2})\mathbf 1_{[0,\mathrm e^{n^2}]}(x).
$$
Then $f$ is decreasing, $g(x)=f(x)x\log(x)$ does not converge to zero when $x\to\infty$ since $g(\mathrm e^{n^2})\geqslant\mathrm e^{-1}$ for every $n\geqslant1$, and $f\leqslant f_0$ hence $f$ is integrable.
I think it's not true, for example $f(x)=1/x$

and we have

$\lim_{x\rightarrow 0}xln(x)(1/x)=+\infty$