Limit of function of 2 variables - can I use polar coordinates?

Question

I need to solve a limit of a f(x,y) (as a part of bigger task), but I'm bad at math. So basically here's this limit:

$$\lim_{x,y\to(0,0)} \frac{y^4}{(x^2+2y^2)\sqrt{x^2+y^2}}$$

I tried to use other methods, but I just can't understand them (like using $\frac{1}{n}, \frac{1}{n}$ as a replacement for $x, y$ and then other sequence, like $\frac{1}{n}, \frac{2}{n}$ to check if the limit is identical, but it just doesn't work here. So I replaced $x, y$ with polar coordinates $x = r\cos{\phi}, y = r\sin{\phi}$ and I have something like this:

$$\lim_{r\to 0} \frac{r^3\sin^4{\phi}}{\sin^2{\phi} + 2\cos^2{\phi}} = 0$$

Can I now safely assume that the limit of the function above (the first one) is equal 0? If not, what else I have to do? Is there a better, or no, simpler method to calculate this limit?

Thanks in advance.

Answer 1

things can't go wrong because the denominator $$\sin^2 \phi + 2\cos^2 \phi \ge 1 + \cos^2 \phi $$ and the numerator $$r^3 \sin^4 \phi \le r^3 \to 0 \text{ as } r \to 0.$$ therefore the limit of the quotient is zero as $(x,y) \to (0, 0)$ in any manner.

Answer 2

I think that unless you get some indeterminated or "problematic" expression with some case under polar coordinates, you can do the coordinates change and deduce the limit is what you get with polar coordinates, and in your case: zero.

There are cases, like with $\;\frac{xy^3}{x^2+y^6}\;$ when taking polar coordinates seem to imply the limit is zero, yet when we approach $\;\theta\to\frac\pi2\;$ we run into in trouble. I'll leave this for you to check.

In the case you wrote, we can even take a direct approach. Since everything's non-negative we use no absolute values, but observe the factors can be assumed to be less than $\;1\;$ as both variables approach zero:

$$\frac{y^4}{(x^2+2y^2)\sqrt{x^2+y^2}}\le\frac{y^4}{(x^2+y^2)^{3/2}}\le\frac{y^4}{|y|^3}=|y|\xrightarrow[(x,y)\to(0,0)]{}0$$

Answer 3

To know that the limit exists, you must show that whatever path you use to approach $(0,0)$ the limit (seen as a one-variable limit with the variable being the distance from $(0,0)$ along that path, gives the same answer. It is (in the 2-variable case) sufficient to consider just paths which near the limit point can be treated as straight lines.

So your trick of going to polar coordinates is absolutely a right thing to do. The different paths are "labelled" by $\phi$ and the one variable that will go to zero is $r$.

You have erred in your algebra: The correct expression is $$\lim_{r\to 0} \frac{r\sin^4{\phi}}{\sin^2{\phi} + 2\cos^2{\phi}} = 0$$ where that last equallity can be justified by noting that the denominator of the factor that depends on $\phi$ is equal to $1+\cos^2\phi$ which is never zero.

So the limit exists, and your answer of zero is correct.