November 30

If f is continuous on [0,1], how to show that

${\displaystyle \lim _{x\to 0^{+))x\int _{x}^{1}{\frac {f(t)}{t))dt=0?}$

I've been asked to compute the integral, which is easily done by choosing f = 0; but how could you show that it's true for all such f? Please assume only a knowledge of the fundamental theorem of calculus, if possible; I'd like to avoid log. —Anonymous Dissident^Talk 12:29, 30 November 2011 (UTC)[reply]

You can use

${\displaystyle \left|\int _{x}^{1}{\frac {f(t)}{t))dt\right|\leq \max |f|\int _{x}^{1}{\frac {1}{t))dt}$

So it's enough to calculate

${\displaystyle \lim _{x\to 0^{+))x\int _{x}^{1}{\frac {1}{t))\,dt.}$

You can either do this by L'Hopital's rule, or, if that's not allowed, write it as

${\displaystyle x\int _{x}^{1}{\frac {1}{t))\,dt={\sqrt {x))\left({\sqrt {x))\int _{x}^{1}{\frac {1}{t))\,dt\right).}$

By taking derivatives, show that the factor in parenthesis is increasing and bounded below, and therefore has a limit as ${\displaystyle x\to 0^{+))$. Sławomir Biały (talk) 12:46, 30 November 2011 (UTC)[reply]

(e.c.) Just use the fact that a continuous function on a closed bounded interval is bounded: for some ${\displaystyle C}$ there holds ${\displaystyle |f(t)|\leq C}$ for all ${\displaystyle t\in [0,1]}$. So you can bound the absolute value of that expression replacing ${\displaystyle f(t)}$ by the constant ${\displaystyle C}$. In this case integrating then get ${\displaystyle -Cx\log x}$, that notoriously converges to ${\displaystyle 0}$ as ${\displaystyle \scriptstyle x\to 0^{+))$; you conclude thanks to the sandwich theorem. --pma 12:53, 30 November 2011 (UTC)[reply]