Some days ago I answered a question that asked to find
$\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha + 1}}}}dt} $
given that $f$ is continuous in $[0,1]$
I proceeded as follows:
$\eqalign{ & t = x\cdot u \cr & dt = x\cdot du \cr} $
So this is produces:
$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} $
I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume
$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} = f\left( 0 \right)\int\limits_1^\infty {\frac{{du}}{{{u^{\alpha + 1}}}}} = \frac{1}{\alpha }f\left( 0 \right)$
This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put
$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} $
And then
$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} } \right| < \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha + 1}}}}du} < \epsilon \frac{{1 - {x^\alpha }}}{\alpha }$
However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?
Would this work?
Let $P$ be the statement that $\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} = \frac{{f\left( 0 \right)}}{\alpha }$
Then $P$ is true if and only if
$ \forall \epsilon > 0\exists \delta > 0$
Such that if $\left| x \right| < \delta $ then $ \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} - \frac{{f\left( 0 \right)}}{\alpha }} \right| < \epsilon $
But then
$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| < $
$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant $
$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < $
And since
$\frac{{1 - {\delta ^\alpha }}}{\alpha } < \frac{1}{\alpha }$ $\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < \frac{\epsilon }{\alpha } < \epsilon $