TITLE: Correct way to prove the limit: $\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha  + 1}}}}dt} $?
QUESTION [5 upvotes]: 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
$$\frac{{1 - {\delta ^\alpha }}}{\alpha } < \frac{1}{\alpha }$$
$$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < \frac{\epsilon }{\alpha } < \epsilon $$

REPLY [3 votes]: You can finish the proof cleanly if you split the integral in your last line to integrate separately over $[0,1/\sqrt x]$ and $[1/\sqrt x,1/x]$.  For $x$ small enough, $|f(xu)-f(0)|<\epsilon\alpha/2$ for $0<u\le 1/\sqrt x$ since $xu\le\sqrt{x}$, and always $|f(xu)-f(0)|\le 2M$ where $M=\max_{0\le u\le 1}|f(u)|$. Then
$$
\left(\int_1^{1/\sqrt{x}}+\int_{1/\sqrt{x}}^{1/x} \right)
\frac{|f(xu)-f(0)|}{u^{\alpha+1}}du \le \frac{\epsilon\alpha}2 \frac{1-x^{\alpha/2}}\alpha + 2M 
\frac{x^{\alpha/2}-x^\alpha}\alpha < \frac{\epsilon}2+ 2M\frac{x^{\alpha/2}}\alpha.
$$
Now if $x$ is small enough, $2Mx^{\alpha/2}<\epsilon\alpha/2$ and the whole integral is less than $\epsilon$. 
So indeed it vanishes in the limit $x\to0$ if $\alpha>0$.