How do you prove the limit $\lim_{(a,b)\rightarrow (0,0))}\frac{x^3y^2}{x^4+y^6}$

Question

$$\lim_{(a,b)\rightarrow (0,0))}\frac{x^3y^2}{x^4+y^6}$$

i tried approaching from y = mx and various polynomials of x which all produced 0. i am told that i should now be using squeeze theorem but im not too sure how to find the right function

Answer 1

if $(x,y)$ approaches $(0,0)$ while $x^3=y^2$, this doesn't tend to 0, so... (sorry, I read $\frac{x^2y^3}{x^4+y^6}$.) You may want to try $X^2=x^4$ and $Y^2=y^6$, and use polar coordinates $X=r\cos\theta$, $Y=r\sin\theta$. You should get a $r^{\alpha}$ estimate, which readily gives the answer...

Answer 2

A little trick for this kind of problems:

Heuristically you can assume that $x$ and $y$ approach $0$ at different rates that can be measured by $x \sim t^{\alpha}$ and $y \sim t^{\beta}$. Under this assumption, the numerator converges to zero at rate $3\alpha + 2\beta$, and the denominator converges at rate $\min{4\alpha , 6\beta}$

If you can find $\alpha$ and $\beta$ such that $3\alpha + 2\beta < 4\alpha$ and $3\alpha + 2\beta < 6\beta$ then you can let $x$ and $y$ approach $0$ accordingly and see that the function does not converge. If you can prove that either $3\alpha + 2\beta > 4\alpha$ or $3\alpha + 2\beta > 6\beta$ then you can prove that the function is convergent by a discussion on both cases.

For example here you can see that $3\alpha + 2\beta < 4\alpha \Leftrightarrow 2\beta < \alpha$ and that $3\alpha + 2\beta < 6\beta \Leftrightarrow \alpha < \frac43 \beta$ so either of them must be false because $\alpha$ cannot be greater the than $2\beta$ and smaller than $\frac43 \beta$ at the same time. So you can pick a number between $\frac43$ and $2$ say $\frac53$ then discuss the following $2$ cases :

1. $|x|>|y|^{\frac53}$, then $x^3 y^2 < x^{3 + \frac65} = x^{4.2}$, and $x^4 + y^6 > x^4$ thus $\frac{x^3y^2}{x^4 + y^6} < x^{0.2}$.
2. $|x| \leqslant |y|^{\frac53}$, then $x^3 y^2 \leqslant y^{5 + 2} = y^{7}$, and $x^4 + y^6 \leqslant y^6$ thus $\frac{x^3y^2}{x^4 + y^6} \leqslant y$.

So in summary $\frac{x^3y^2}{x^4 + y^6} \leqslant \max{\{x^{0.2},y\}} < \left(x^2 + y^2\right)^{0.1}$, thus converges to $0$.