# Trying to to solve a limit by change of variables: legit or not.

I was trying to solve this limit: $$\lim_{x^2+y^2\to+\infty\\\; x\ge0,\;y\ge0}\frac{x^2y^3+\sin(x^2y)}{1+x^4+|y|^7}$$
1.$$\frac{x^2y^3+\sin(x^2y)}{1+x^4+|y|^7} \le \frac{x^2y^3+1}{1+x^4+|y|^7}$$ $$\qquad$$ 2. $$\begin{cases} x^4=u^2\rightarrow |x|=|u|^{1/2} \\ |y|^7=v^2 \rightarrow|y|=v^{2/7} \end{cases}$$ $$\qquad$$ 3.$$\; \Rightarrow \frac{1+|u|\cdot v^{6/7}}{1+u^2+v^2}$$

Now, my problem is understand how {$$x\ge0, y\ge0$$} transforms with respect to the change of variables I did.
In case the transformation keeps {$$u\ge0, v\ge0$$}, then a second change in polar coordinates will solve the limit.
Also, I may have reason to believe {$$x\ge0, y\ge0$$}$$\ \to \$${$$(u,v) \in \Bbb R^2$$}, and again polar coordinates will do the job.

So, is that change of variables legit in order to solve the limit? And more generally, how will {$$x\ge0, y\ge0$$} transform?

• You don't need absolute value signs anywhere.
– zhw.
May 6, 2020 at 21:35
• @zhw Why are you stating that? If weren't absolute value at all, the transformation keeps {$u\ge0, v\ge0$}? May 8, 2020 at 16:48
• Your limit specifies $x\ge 0, y\ge 0.$
– zhw.
May 8, 2020 at 17:24
• @zhw And what can I conclude with that? Maybe {$x\ge0, y\ge0$}$\rightarrow${$(u,v)\in \Bbb R^2)$} May 10, 2020 at 15:18
• You can conclude you don't need absolute value signs anywhere.
– zhw.
May 10, 2020 at 15:27