When can we say $ \lim_{x \to 0} \frac{1}{\sin x} \int_{0}^{\sin x} {x}^2 f(x) dx = 0$? I encountered this problem while trying to solve a larger problem involving oscillation in a nonlinear system.
What conditions need to be applied on $f(x)$ such that,
$$ \lim_{x \to 0} \frac{1}{\sin x} \int_{0}^{\sin x} {u}^2 f(u) \:\textrm{d}u = 0  \:\:\textrm{holds.}$$
We are given that the function $f(x)$ is bounded, monotonically decreasing, and non-negative. Is this enough to guarantee that the condition above holds?
Can we relax the bounded condition and still have it be true?
 A: Since we have boundedness, $0 \leq f(x) \leq C$, it will be true.  (You don't need $f(x)$ to be non-negative, just bounded below by some number).  Denote $L$ as the limit, then 
$L \leq \lim_{x\rightarrow 0} \frac{1}{\sin x}\int_{0}^{\sin x}Ct^{2} dt=\lim_{x\rightarrow 0} \frac{C(\sin x)^{3}}{3\sin x} = \lim_{x\rightarrow 0}\frac{C (\sin x)^{2}}{3}=0$.
The lowerbound is proven similarly.
To answer the updated question: 
Using the same argument, if we relax the boundedness of f(x), as long as for some $\epsilon>0$, $x^{2-\epsilon}f(x)$ is bounded in an open interval containing $0$, then the limit equals $0$.
If $x^{2}f(x)$ is not bounded then we can construct a counterexample.
Say $f(x) = \frac{1}{x^{2.5}}$, so that the integral in the limit exists.
Then $$\lim_{x\rightarrow 0^{+}} \frac{1}{\sin x}\int_{0}^{\sin x}\frac{t^{2}}{t^{2.5}} dt=\lim_{x \rightarrow 0}  \frac{2\sqrt{\sin x}}{\sin x} = \lim_{x \rightarrow 0^{+}}  \frac{2}{\sqrt{\sin x}} = \infty$$
A: As other have said, if $f$ is non-negative and bounded above, it will work. Another way to see this is L'hopital's rule
