How to (dis)prove the existence of this limit? Let $f:(0,1] \to \mathbb R$ be a bounded integrable function. How to (dis)prove the existence of this limit? 
$$\lim_{x \to 0+} \frac{1}{|\ln(x)|} \int_x^1 \frac{f(t)}{t} \mathrm dt $$
If the limit doesn't exist, what kind of assumption(s) should be given to guarantee the limit? Also, is there a name or kind for this limit/integral?
Thank you. 
 A: Let $J(x) = \displaystyle \int_x^1 \frac{f(t)}{t}\ dt$.
Note that if $|f(x)| \le B$ for all $x$, $\left|J(x)\right| \le - B \ln(x)$, so the lim sup and lim inf are finite.  However, consider $\displaystyle f(x) = (-1)^{\lfloor \ln(-\ln(x)) \rfloor}$, i.e. for $\exp(-e^n) < x \le \exp(-e^{n-1})$, $f(x) = 1$ if $n$ is odd and $-1$ if $n$ is even.  Then 
$$(-1)^{n-1} J(\exp(-e^n)) > \int_{\exp(-e^n)}^{\exp(-e^{n-1})} \dfrac{dt}{t} - \int_{\exp(-e^{n-1})}^1 \dfrac{dt}{t} = (1 - 2/e) e^n $$
so $\lim_{x \to 0+} J(x)/|\ln(x)|$ does not exist.
EDIT: It may be useful to make a change of variables: if $g(s) = f(e^{-s})$, 
$$\lim_{x \to 0+} \frac{J(x)}{|\ln(x)|} = \lim_{y \to \infty} \frac{1}{y} \int_0^y g(s)\ ds$$ 
(each limit existing if and only if the other does).  
Note that if $\int_{n}^{n+1} |g(s)|\ ds$ is bounded, the limit exists and is $L$ if and only if the series $\sum_{k=1}^\infty a_k$ is Cesàro summable to $L$, where $a_1 = \int_0^1 g(s)\ ds$ and $a_{k} = \int_{k-1}^k g(s)\ ds - \int_{k-2}^{k-1} g(s)\ ds$ for $k \ge 2$.
A: If $\displaystyle\lim_{x\to0^+}f(x)=L$, then $\displaystyle\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t=L$.
Choose an $\epsilon>0$ and find $\delta>0$ so that if $0<x<\delta$, then $|f(x)-L|<\epsilon$.
Then
$$
\begin{align}
&\left|L-\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t\right|\\
&=\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\
&=\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^\delta\frac{f(t)-L}{t}\,\mathrm{d}t
+\lim_{x\to0^+}\frac1{|\log(x)|}\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\
&\le\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^\delta\frac{f(t)-L}{t}\,\mathrm{d}t\right|
+\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\
&\le\epsilon\lim_{x\to0^+}\left|\frac{\log(x)-\log(\delta)}{\log(x)}\right|
+\lim_{x\to0^+}\frac1{|\log(x)|}\left|\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\
&=\epsilon+0
\end{align}
$$
Therefore,
$$
\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t=L
$$
