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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?

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    You can check out my user page, it has a lot of similar questions that might be of interest to you.2012-04-09

2 Answers 2

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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, $L=\ell$.

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    First we work in the case on which $\ell=0$, , using $g=f-\ell$, then we try to generalize.2012-04-08
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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!

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    Fair enough. Let me instead say that the version of l'Hopital that we often use in practice - that is, just take derivatives of numerator and denominator and see what happens - is not valid unless the original limit has an indeterminate form.2012-04-09