Exercise of fundamental theorem of calculus Let $f$ is continuous on $[0,1]$, then 
$\lim\limits_{x\rightarrow0 ^+} x^2\int_{x}^{1} \frac{f(t)}{t^3} dt~ = ~?  $
The problem is form chapter of FTC, but I have no idea of this problem.
And i think that the solution is depend on the function $f$ ?
 A: Use L'Hôpital's rule (and FTC):
$$\lim\limits_{x\rightarrow0 ^+} x^2\int_{x}^{1} \frac{f(t)}{t^3} dt=
\lim\limits_{x\rightarrow0 ^+} \frac{\int_{x}^{1} \frac{f(t)}{t^3} dt}{1/x^2}\overset{\text{L'Hôpital}}{=}
\lim\limits_{x\rightarrow0 ^+} \frac{-\frac{f(x)}{x^3}}{-2x^{-3}}=\frac{f(0)}{2}.$$
P.S. It is not necessary to check that numerator goes to infinity. See  How does the L.Hopital rule work when numerator $\neq$ $\infty$. Anyway, you can also say that if numerator does not go to infinity then necessarily $f(0)=0$ (otherwise $\frac{f(t)}{t^3}\sim \frac{f(0)}{t^3}$ whose improper integral in a right neighborhood of $0$ is not convergent) and the limit is easily seen to be $0$. So the result $f(0)/2$ holds also in this case.
A: Here is another take on this without the use of L'Hospital's Rule. We have
\begin{align}
g(x) &= x^{2}\int_{x}^{1}\frac{f(t)}{t^{3}}\,dt\notag\\
&= x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt + x^{2}f(0)\int_{x}^{1}\frac{dt}{t^{3}}\notag\\
&= x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt + x^{2}f(0)\left(\frac{1}{2x^{2}} - \frac{1}{2}\right)\notag\\
&= \frac{f(0)}{2} - \frac{x^{2}f(0)}{2} + x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\notag
\end{align}
As $x \to 0^{+}$ we can see that the first term above remains constant, the second term goes to $0$ and we shall prove below that the third term also goes to $0$ and hence the desired limit is $f(0)/2$.
Let $\epsilon > 0$ be given. Then by continuity of $f$ at $0$ it follows that there is an $h > 0$ such that $$|f(t) - f(0)| < \epsilon$$ for all $t$ with $0 \leq t \leq h$ and hence if $0 < x < h$ then
\begin{align}
h(x) &= \left|x^{2}\int_{x}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|\notag\\
&\leq \left|x^{2}\int_{x}^{h}\frac{f(t) - f(0)}{t^{3}}\,dt\right| + \left|x^{2}\int_{h}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|\notag\\
&< \epsilon x^{2}\int_{x}^{h}\frac{dt}{t^{3}} + x^{2}M(h)\notag\\
&= \epsilon x^{2}\left(\frac{1}{2x^{2}} - \frac{1}{2h^{2}}\right) + x^{2}M(h)\notag\\
&< \frac{\epsilon}{2} + x^{2}M(h)\notag
\end{align}
where $$M(h) = \left|\int_{h}^{1}\frac{f(t) - f(0)}{t^{3}}\,dt\right|$$ is independent of $x$. By the above equation it is clear that $$\limsup_{x \to 0^{+}}h(x) \leq \frac{\epsilon}{2}$$ Since $\epsilon > 0$ was arbitrary it follows that $\lim_{x \to 0^{+}}h(x) = 0$ and we are done.
