Limit of a bivariate fraction

Question

Does the limit as $(x, y)$ approaches $(0, 0)$ of $\frac{x^4y^5}{x^4+2y^2}$ exist?

I tried approaching $(0,0)$ from the x-axis, y-axis and $y=x^2$. They all equal to $0$.

Answer 1

Use polar substitution $x=r\cos{\theta}$ and $y=r\sin{\theta}$. Then, $\frac{x^4y^5}{x^4+2y^2}=\frac{r^7\cos^4{\theta}\sin^5{\theta}}{r^2\cos^4{\theta}+2\sin^2{\theta}}$ goes to 0 when $r\to 0$

Answer 2

A good definition of a limit$L$ of $f(x,y)$ as the two variables $x,y$ approach zero is that for any given positive real $\delta$ there exists some $\epsilon$ such that $$ (0<|x| < \epsilon) \wedge (0<|y| < \epsilon) \implies |L - f(x,y)| < \delta $$ In the case of $\frac{x^5y^4}{x^4+2y^2}$ and $L = 0$, you can take $\epsilon = \min( \frac{\delta}2, \frac12)$. To show this, look at two cases: $\delta\geq 1$, and $\delta < 1$. Without loss of generallity we can take $x \geq 0$ so that $f(x,y) \geq 0$.

For $\delta\geq 1$, if $x^2=\sqrt{2}y$ then

$$\frac{x^5y^4}{x^4+2y^2} =\frac{x^{13}}{2x^4}=\frac{x^9}{8} < \frac{\delta^9}{2^{12}}<\frac{1^8}{2^{12}}\delta < \delta $$ If $x^2>\sqrt{2}y$ $$\frac{x^5y^4}{x^4+2y^2} =xy^4-\frac{2xy^6}{x^4+2y^2}\leq xy^4 \frac{x^9}{4} < \frac{\delta^9}{2^{11}} < \delta $$ And so forth.

However, it is much easier if you can utilize various theorems about limits.