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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?

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

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