$f$ a continuous function in [0,1], Show: $\lim_{x \to 0}, x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}}$, for $\alpha >0$ Let $f$ be a continuous function in [0,1] and $\alpha >0$, I'd love your help with finding the following limit:
 $\lim_{x \to 0},  x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}}$.
First I tried to bound the function since it is continues in a closed interval
$\lim_{x \to 0},  |x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}} |\leq  \lim_{x \to 0},|  x^{\alpha}\int_{x}^{1}\frac{M}{t^{\alpha+1}}|$ but I get a number depends on $\alpha$ and $M$, and it won't do. So, I assume that the integral is always diverges, since $f$ is continues and it is divided by $t$ with exponent bigger than one, so it is 0 multiplied by $\infty$, maybe we can use L'Hopital somehow?
Thanks!
 A: Let $\varepsilon>0$ and $\delta$ such that if $|x|\leq\delta$ then $|f(x)-f(0)|\leq\varepsilon$. We have for $0<x<\delta$:
\begin{align*}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right|
&\leq x^{\alpha}\int_x^{\delta}\frac{\varepsilon}{t^{\alpha+1}}dt+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\
&=x^{\alpha}\varepsilon \frac{-1}{\alpha}(\delta^{-\alpha}-x^{-\alpha})+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\
&=\frac{(1-x^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\
&\leq \frac{(1+|x|^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+|x|^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt,
\end{align*}
so $\limsup_{x\to 0}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right|\leq \varepsilon$ and $\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)-f(0)}{t^{\alpha+1}}dt=0$. 
Since 
$$x^{\alpha}\int_x^1\frac{dt}{t^{\alpha+1}}=-\frac 1{\alpha}(1-x^{-\alpha})x^{\alpha}=\frac 1{\alpha}(1-x^\alpha),$$
so finally 
$$\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)}{t^{\alpha+1}}dt=\frac{f(0)}{\alpha}.$$
A: You said it yourself in the question: use L'Hospital's Rule applied to $$\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}$$
The numerator and denominator each approach $\infty$, given your conditions on $f$. 

EDIT: The denominator approaches $\infty$ simply because $\alpha$ is positive. 
If $f(0)$ is nonzero, then $f$ is bounded below by some positive $\epsilon$ in a neighborhood of $0$. So the integral is bounded below by $\epsilon\int_x^\delta\frac{1}{t^{\alpha+1}}\,dt+\int_{\delta}^1\frac{f(t)}{t^{\alpha+1}}\,dt$ which diverges to infinity as $x$ approaches $0^+$, since the power of $t$ is greater than $1$.
(And if $f(0)=0$ the numerator might not approach $\infty$. But it still approaches something since $f$ is integrable on $[0,1]$ and $t^{\alpha+1}$ is monotonic. Let's call it $L$. Then L'Hospital's Rule is not needed - the OP's limit is $0\cdot L$, or just $0$. This is consistent with the formula given below when $f(0)\neq0$.)

The result is 
$$\begin{align}\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}&=\lim_{x\to0^+}\frac{\frac{d}{dx}\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}1/x^{\alpha}}\\
&=\lim_{x\to0^+}\frac{-\frac{d}{dx}\int_1^x\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}x^{-\alpha}}\\
&=\lim_{x\to0^+}\frac{-\frac{f(x)}{x^{\alpha+1}}}{-\alpha x^{-\alpha-1}}\\
&=
\lim_{x\to0^+}\frac{f(x)}{\alpha}\\
&=\frac{f(0)}{\alpha}
\end{align}$$
