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Here are two examples:

$$ \lim_{(x,y)\to(0,0)} \frac{3xy}{x^2+3y^2} $$

and

$$ \lim_{(x,y)\to(0,0)} \frac{5xy^2}{x^2+2y^2} $$

My question is, what process should I go through in determining which approach to use on these problems? I already know how to solve simpler limits that require factorizations or multiplying by a conjugate. The first example is solved by showing that the value of the limit along two different paths is different, so the limit therefore does not exist, and the second example is solved using the squeeze theorem. What approach should be taken in order to determine which of these methods should be used, or other methods (including evaluating the limit of a generic path such as $y=mx$)?

Thanks!

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I'm always a fan of using polar coordinates when possible. One easy usage of this is if the power of $r$ in the numerator exceeds the power of $r$ in the denominator then the limit is $0$ as $(x,y)\to 0\iff r\to 0$. (here $r$ is the polar coordinate radius).

For example, in the second limit: the power of $x$ and $y$ in the product adds to $3$, whereas in the denominator the powers of $x$ and $y$ are both $2$. So when we change to polar coordinates and factor out the $r$ terms we get a $r^3$ over $r^2$ type limit. That is, $$\lim_{(x,y)\to 0}\frac{5xy^2}{x^2+2y^2}=\lim_{r\to 0}\frac{r^3(5\cos(\theta)\sin^2(\theta))}{r^2(\cos^2(\theta)+2\sin^2(\theta))}=0$$ since $\lim_{r\to 0}\frac{r^3}{r^2}=0$. Be careful though, as when using polar coordinates the limit must exist independently of $\theta$.

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For the first one, both the denominator and numerator go to zero whenever $(x,y)\rightarrow(0,0)$, but they have the same degree, which is of second degree, so it suggests testing by $y=mx$.

For the second one, both the denominator and numerator go to zero as well, but the numerator has the third degree, the denominator has only the second degree, so it suggests that the numerator will go faster than denominator, so the limit should be zero, so trying Squeeze Theorem gets through.

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both these follow from $$ x = r \, \sqrt A \; \cos \theta \; , \; \; y = r \sin \theta $$ and see whether letting $r$ go to zero forces a limit

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  • $\begingroup$ What is $A$ here when you switch to polar coordinates? $\endgroup$ – Clement C. Apr 29 '18 at 3:41

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