# $\lim\limits_{y \rightarrow 0}3y\int_{y^2}^{y}\frac{x^2\sin(x^2y^3)}{x^2+\cos(x^2y^3)}dx\quad x,y \in R$

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 $$a, then $$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...

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 $$3|y|\cdot (2/\cos 1) \to 0.$$

• Ok this seams to be good, but, i don't understand how to you get such precision for $\frac{2}{\cos1}$. The only thing i have is $\mid\frac{x^2\sin(x^2y^3)}{x^2+\cos(y^3x^2)}\mid \leq \frac{1}{1-(y^6x^4)}<1$ because near $0$ we have $\cos(h)>1-h^2$ ? – Dicordi Jun 12 at 16:39
• $y^3x^2$ lies in the interval $[-1,1],$ hence $\cos y^3x^2\ge \cos 1$ in this interval. – zhw. Jun 12 at 17:19

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!!

• I think that since $y$ is also in the integration limits it's not valid to simply flip the integral and limit. – J_P Jun 12 at 16:05
• Yes exact, but in this case I apply the theorem I wrote, and because $\mid y \mid <1$,then $y-y^2$ is not $(0)$, so... I don't really understand what are you trying to tell me here – Dicordi Jun 12 at 16:06
• Your theorem only applies when the limits of integration are constant. – J_P Jun 12 at 16:07
• Ok thank you $J_P$, so know I really don't know how to compute this limit... – Dicordi Jun 12 at 16:09
• @Dicordi Since $|y|<1$ then the condition on the integration limits $a<b$ in the theorem are not guaranteed; there will be cases when $a>b$ and the theorem does not clearly seem to accomodate that possibility. – Majura Selekwa Jun 12 at 16:21

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 $$|f(x,y)| for some constant $$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$$.

• Why do you have the integral between $-\mid y\mid$ and $\mid y \mid$ in the middle ? – Dicordi Jun 12 at 16:28
• Because $|y|\geq y,y^2\geq-|y|$ and so when you're integrating from $y^2$ to $y$ you get even more if you integrate from $-|y|$ to $|y|$ and the inequality is OK. I only did it that way because if $y$ is perhaps negative then $y<y^2$ and I don't want to bother distinguishing between cases. – J_P Jun 12 at 16:50