Find: $\lim\limits_{x\to 0}{x^{\alpha}\int_{x}^{1}{f(t)\over t^{\alpha +1}}dt}$. Let $f$ be continuous on $[0,1]$, and let $\alpha>0$. Find: $\lim\limits_{x\to 0}{x^{\alpha}\int_{x}^{1}{f(t)\over t^{\alpha +1}}dt}$. I tried integration by parts, but I am not sure if $f$ is integrable and to what extent. It also gets really messy. Besides, I am not sure if I am to express the limit using $f$, or to arrive at an actual number. It would be nice if you could take a look.
Using other questions I got: Let us denote $G(x)=\int_{x}^{1}{f(t)\over t^{\alpha +1}}dt$, so I am looking for: $\lim\limits_{x\to 0}{x^{\alpha}G(x)}=\lim\limits_{x\to 0}{G(x)\over {1\over x^{\alpha}}}$. If $g(t)=\int{f(t)\over t^{\alpha +1}}dt$, Then: $G(x)=g(1)-g(x)$ which means: $G'(x)=-g'(x)=-{f(x)\over x^{\alpha +1}}$. Let us use L'Hôpital's rule: $\lim\limits_{x\to 0}{G(x)\over {1\over x^{\alpha}}}=\lim\limits_{x\to 0}{{G'(x)=-{f(x)\over x^{\alpha +1}}}\over {-\alpha\over x^{\alpha+1}}}=\lim\limits_{x\to 0}{f(x)\over \alpha}={f(0)\over \alpha}$.
 A: Too long for a comment: Since $f$ is uniformly continuous on $[0,1]$, it is bounded on the interval, so suppose $|f| \leq M$. Now, we deal with this integral:
\begin{align}
&\lim \limits_{x \to 0} x^{\alpha} \int_x^1 \frac{M}{t^{\alpha + 1}} \, dt\\
&\lim \limits_{x \to 0} M x^{\alpha} \left[ - \frac{1}{\alpha t^{\alpha}} \right]_x^1\\
&\lim \limits_{x \to 0} \frac{M}{\alpha} - \frac{M}{\alpha} x^{\alpha}\\
&= \frac{M}{\alpha}
\end{align}
So the integral is bounded by that.
A: Choose $\epsilon > 0$. Since $f$ is continuous, we can take $\delta \leq 1$ such that $$m_{\delta}=f(0)-\epsilon < f(t) < f(0)+\epsilon=M_{\delta}$$ on $[0,\delta]$. Then we have
\begin{align}\lim_{x \to 0} x^\alpha \int_x^1 \frac{f(t)}{t^{\alpha+1}} \, dt&=\lim_{x \to 0} x^{\alpha} \int_x^\delta \frac{f(t)}{t^{1+\alpha}}\, dt+\lim_{x \to 0}x^\alpha\int_{\delta}^1 \frac{f(t)}{t^{1+\alpha}}\, dt \\
&\leq\lim_{x \to 0} x^{\alpha} \int_x^\delta \frac{M_{\delta}}{t^{1+\alpha}}\, dt + 0\\
&=\lim_{x \to 0} x^{\alpha}\left(\frac{M_{\delta}}{-\alpha \delta^\alpha}- \frac{M_{\delta}}{-\alpha x^\alpha}\right) \\
&=\frac{M_{\delta}}{\alpha}
\end{align}
and similarly $\lim_{x \to 0} x^\alpha \int_x^1 \frac{f(t)}{t^{\alpha+1}} \, dt \geq \frac{m_{\delta}}{\alpha}$.
Thus the limit is bounded below by $\frac{f(0)}{\alpha}-\frac{\epsilon}{\alpha}$ and above by $\frac{f(0)}{\alpha}+\frac{\epsilon}{\alpha}$. Since $\epsilon$ was arbitrary, it follows that the limit must be equal to $\frac{f(0)}{\alpha}$.
(As a side note, this proof implies that $f$ doesn't have to be continuous on all of $[0,1]$; it suffices if it's continuous at $0$ and integrable on $[0,1]$.)
