Continuity of a rational function I have to evaluate the continuity of a two functions at a couple of different points and I am a bit stuck. Here are the two functions:
$f(x,y) = 0$ if $(x,y)=(0,0)$, $f(x,y) = \frac{x^3y^2}{3x^4+2y^2}$ if $(x,y) \neq (0,0)$. I want to evaluate the continuity of this one at $(0,0)$ and $(1,0)$.
I also have $f(x,y) = 0$ if $(x,y)=(0,0)$, $f(x,y) = \frac{y^3x}{x^2+y^6}$ if $(x,y) \neq (0,0)$. I want to evaluate the continuity of this one at $(0,0)$ and $(0,1)$. 
Okay, so for both of these functions at $(0,0)$ the denominator is zero along $3x^4+2y^2$ and $x^2+y^6$, respectively, so I cannot simply evaluate the limit of a sequence approaching points along this line to determine the limit. Everywhere else however, including $(1,0)$ the limit exists and is hence continuous.
In my class, we did two similar examples, but both denominators were of the form $x^2+y^2$. It was easy to show that for one example, you get a different limit for various sequences approaching the origin, hence the limit DNE. For the other example, we proved a limit existed by using the squeeze theorem. But both ways seemed more to be like tricks to me. How am I supposed to know what to do here without any experience?
 A: By experience this kind of limit always seems to fall (if it falls at all) to a curve of the shape
$$ (t^a,t^b) \to (0,0) \text{ for } t\to 0 $$
where $f(t^a,t^b)$ has a nonzero limit. Let's try how that works out in the first case. We get
$$ \frac{x^3y^2}{3x^4+2y^2}  = \frac{t^{3a+2b}}{3t^{4a}+2t^{2b}} $$
This will go to zero iff the dominant exponent in the numerator -- that is, $3a+2b$ -- is larger than the dominant exponent in the denominator -- that is, $\min(4a,2b)$ -- so in order to be a counterexample our $a,b$ need to satisfy
$$ 3a+2b \le \min(4a,2b) \iff \begin{cases} 3a+2b \le 4a \\ 3a+2b \le 2b \end{cases} $$
But the second of these inequalities is obviously impossible (because $a$ has to be positive), so we can't prove with a curve of this form that the limit doesn't exist. This makes us suspect that the limit is 0, but we need to prove this later.
Before that, though, let's try the same technique on the second case. Here we have
$$ \frac{y^3x}{x^2+y^6} = \frac{t^{a+3b}}{t^{2a}+t^{6b}} $$
and so we want
$$ a+3b \le \min(2a,6b) \iff \begin{cases} a+3b \le 2a \\ a+3b \le 6b \end{cases} \iff \begin{cases} 3b \le a \\ a \le 3b \end{cases} $$
This is satisfiable, just barely, by setting $(a,b)=(3,1)$, so
$$ \text{for }(x,y)=(t^3,t)\text{ we have }\frac{y^3x}{x^2+y^6}=\frac{t^6}{t^6+t^6}=\frac 12 $$

Back to the first function. When $(x,y)\to 0$ we certainly also has $3x^4+2y^2\to 0$, so let's set $u=3x^4+2y^2$ and see what happens to the fraction.
We have $3x^4<u$ and therefore $x < (u/3)^{1/4}$. Similarly $2y^2<u$ and therefore $y^2 < u/2$. Thus,
$$ \frac{x^3y^2}{3x^4+2y^2} = \frac{x^3y^2}{u} < \frac{(u/3)^{3/4}(u/2)}{u} = \frac1{2\cdot 3^{3/4}} u^{3/4} $$
which certainly goes to $0$ when $u\to 0$. So the first function is really continuous at $(0,0)$.
