$\lim\limits_{y \rightarrow 0}3y\int_{y^2}^{y}\frac{x^2\sin(x^2y^3)}{x^2+\cos(x^2y^3)}dx\quad x,y \in R$ I would like to compute the following limit : 
$$\lim\limits_{y\rightarrow 0}{}g(y)=\lim\limits_{y \rightarrow 0}3y\int_{y^2}^{y}\frac{x^2\sin(x^2y^3)}{x^2+\cos(x^2y^3)}dx\quad x,y \in R, \quad \mid y \mid <1$$
My attempt : 
We can use this theorem : 
If $f :[a,b] \times I \rightarrow R$ is continuous ($I$ is open) and $\frac{\partial f}{\partial y}$ exists and is continuous, then , let $a<b$, then
$$g(y):=\int_{a}^{b}f(x,y)dx$$ is $C^1(I)$ and
$$g'(y):=\int_{a}^{b}\frac{\partial f}{\partial y}(x,y)dx$$
It means that $\lim\limits_{y\rightarrow 0}g(y)=g(0)$
So here, since $\mid y \mid <1$, we get 
$$\lim\limits_{y\rightarrow 0}{}g(y)= 0$$ ? I am not sure about that...
 A: By direct inspection the limit seems to be 0.
Set $$\lim\limits_{y\rightarrow 0}{}g(y)=\int_{y^2}^{y}\lim\limits_{y \rightarrow 0}\left[\frac{3yx^2\sin(x^2y^3)}{x^2+\cos(x^2y^3)}\right]dx\quad x,y \in R$$
which turns out to be
$$\int _{y^2}^{y}0dx=0$$
I could be wrong!!
A: Suppose $|y|<1.$ Then $y,y^2\in [-1,1].$ Thus the absolute value of the integral is bounded above by the length of $[-1,1]$ times the maximum of the integrand in absolute value over this interval. This is no more than
$$2\cdot \frac{1}{\cos 1}.$$
Thus the total expression has absolute value no more than $3|y|\cdot (2/\cos 1) \to 0.$
A: The function of two variables
$$
f(x,y)=\frac{3x^2y\sin(x^2y^3)}{x^2+\cos(x^2y^3)}
$$
is continuous at the origin $(0,0)$, so $|f(x,y)|<M$ for some constant $M$ in a neighbourhood of the origin. In the integral, $x$ runs from $y^2$ to $y$, so in particular, $|x|\leq|y|$. Then when $y$ is small enough, we have
$$
\left\vert\int_{y^2}^{y}f(x,y)\mathrm{d}x\right\vert\leq\int_{-|y|}^{|y|}|f(x,y)|\mathrm{d}x<\int_{-|y|}^{|y|}M\mathrm{d}x=2M|y|
$$
But $\lim_{y\rightarrow 0}2M|y|=0$.
