Is it true $$\lim_{x\to+0}x^{p}\int^{x}_1w^{-p}f\left(w\right)dw=0,$$ where $f\in C^{0}(\mathbb{R})$, and $p>0$?
In a case $f$ is an analytic, this relation can be easily proved by decomposing $f$ into Taylor series. Another way is integration by parts, but this also requires the function $f$ to be at least $\left\lfloor p\right\rfloor$ times differentiable. But how to prove if $f$ is just bounded?
Since $f$ is bounded, there is an $M$ such that $|f(x)| \le M$ for all $x\in[0,1]$. Then $$ \left|x^p\int_x^1 w^{-p}f(w)dw\right|\le x^p\int_x^1 \left|w^{-p}f(w)\right|dw \le x^p\int_x^1 M w^{-p} dw . $$ This integral can be evaluated, giving $$ x^p\int_x^1 M w^{-p} dw =\begin{cases}Mx^p \frac{1-x^{1-p}}{1-p} & 0<p<1 \\ Mx \ln(x) & p = 1 \\ Mx\frac{1-x^{p-1}}{p-1} & p>1\end{cases}. $$ Since all three of those clearly go to zero as $x \rightarrow 0^+$, and they bound the magnitude of the original integral, it must also go to zero.