I'm solving this problem and I guess it shouldn't be too hard. Since $f$ is continuous it is bounded, so one has
$\left| {x\int\limits_x^1 {\frac{{f\left( t \right)}}{t}dt} } \right| \leq x\int\limits_x^1 {\left| {\frac{{f\left( t \right)}}{t}} \right|dt} \leqslant Mx\int\limits_x^1 {\frac{{dt}}{t}} = - Mx\log x \to 0$
Where $M=\operatorname{sup}\{|f(x)|:x\in[0,1]\}$
I'm not 100% certain on this, so I want a better, clearer approach.
Then, there is a second problem, similar, which is:
If $f$ is integrable on $[0,1]$ and $\exists\lim\limits_{x\to0}f(x)=L$, find
$\ell = \mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} $
ADD: The second might follow from the first since
\mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} =\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) - xf\left( 1 \right) + x\int\limits_x^1 {\frac{{f'\left( t \right)}} {t}dt}
= L + \mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f'\left( t \right)}}{t}dt}
So, what can I sat about f'(t) given $f(t)$ is integrable on $[0,1]$ that will allow me to apply the first case to the last limit?