If $f$ is continuous in $[0,1]$, find $\lim\limits_{x \to 0^{+}}x\int\limits_x^1 \frac{f(t)}t dt$ 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?
 A: If $f$ is continuous let's use L'Hopital rule
$$
\lim\limits_{x\to+0}x\int\limits_{x}^{1}\frac{f(t)}{t}dt=
\lim\limits_{x\to+0}\frac{\int\limits_{x}^{1}\frac{f(t)}{t}dt}{x^{-1}}=
\lim\limits_{x\to+0}\frac{-\int\limits_{1}^{x}\frac{f(t)}{t}dt}{x^{-1}}=
\lim\limits_{x\to+0}\frac{-\frac{f(x)}{x}}{-x^{-2}}=
\lim\limits_{x\to+0}xf(x)=0
$$
$$
\lim\limits_{x\to+0}x\int\limits_{x}^{1}\frac{f(t)}{t^2}dt=
\lim\limits_{x\to+0}\frac{\int\limits_{x}^{1}\frac{f(t)}{t^2}dt}{x^{-1}}=
\lim\limits_{x\to+0}\frac{-\int\limits_{1}^{x}\frac{f(t)}{t^2}dt}{x^{-1}}=
\lim\limits_{x\to+0}\frac{-\frac{f(x)}{x^2}}{-x^{-2}}=
\lim\limits_{x\to+0}f(x)
$$
P.S. Big thanks to David Mitra, he pointed out that requirement for integrals to be divergent is unnecessary!
A: For the first problem, your approach is fine (but the first inequality maybe be an equality when $f$ is non-negative). For the second, denote $L:=\lim_{x\to 0}f(x)$. Fix $\varepsilon>0$. We can find $\delta>0$ such that if $0\leq x\leq \delta$ then $|f(x)-L|\leq \varepsilon$, so for $0\leq x\leq \delta$:
$$x\int_x^1\frac{f(t)}{t^2}dt=x\int_x^1\frac{f(t)-L}{t^2}dt+Lx\int_x^1\frac{dt}{t^2}=x\int_x^1\frac{f(t)-L}{t^2}dt+L\left(\frac 1x-1\right)x$$
hence 
\begin{align*}\left|x\int_x^1\frac{f(t)}{t^2}dt-L\right|&\leq x\int_x^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\
&=x\int_x^\delta\frac{|f(t)-L|}{t^2}dt+ x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\
&\leq x\int_x^\delta\frac{\varepsilon}{t^2}dt+ x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\
&=\varepsilon x\left(\frac 1x-\frac 1{\delta}\right)+x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\
&=\varepsilon-\frac{\varepsilon}{\delta}x+x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|
\end{align*}
so 
$$\limsup_{x\to 0^+}\left|x\int_x^1\frac{f(t)}{t^2}dt-L\right|\leq \varepsilon$$
and since $\varepsilon$ was arbitrary, $L=\ell$.
