Values of $a$ for which $g(x,y) = \frac{|x^a y|}{|x|^5+y^2}$ tends to zero when $(x,y) \to 0$. Obviously, if the limit exists it must be equal to zero, as
$$g(x,0) = \frac{0}{y^2} = 0,$$
independently of $a$.
We have
$$|g(x,y)| = \left|\frac{x^a y}{|x|^5+y^2}\right|\leq \left|\frac{x^a y}{x^5}\right|$$
which tends to zero iff $a \geq 5$. So, am I correct in assuming that the answer would be just that? ($a \geq 5$.) Or is this result just down to the way I set up the upper bound for $g$, i.e., $g\leq |x^ay/x^5|$?
 A: You've got the right answer (That the function tends to $0$ if and only if $a \geq 5$), but you've only proved half of it: that the limit is $0$ for all values $a \geq 5$. It's still possible that there are values of $a$ less than $5$ where it converges (which your method tells us nothing about) so to complete your answer you need to show that no value of $a<5$ has the function converge.
If $a<5$, consider the curve $y = x^{\frac{5-a}{2}}$. On this curve:
$$|g(x, x^{\frac{5-a}{2}})| = \frac{|x^\frac{5+a}{2}|}{|x^5|+y^2} \geq \frac{|x^\frac{5+a}{2}|}{|x^5|} = |x^{\frac{a-5}{2}}| \rightarrow \infty $$
as $(x,y) \rightarrow 0$. Since the function doesn't tend to $0$ on this curve, it can't tend to $0$ on the entire plane.
A: First, assuming we're working in real numbers, we have a problem since $x^a$ is not, in general, defined as a real number when $x<0$. There's two ways we can fix this: either impose the condition $x\geq 0$ or redefine the function as
$$f(x,y)=\frac{|x|^a|y|}{|x|^5+y^2} $$
I'll take the second approach (the answer is the same for both, though). As you said, if the limit exists, it has to be $0$. Consider the path $(x,y)=(t,t^{5/2})$ with $t>0$:
$$f\left(t,t^{5/2}\right)=\frac{t^{a+5/2}}{t^5+t^5}=\frac 12t^{a-5/2} $$
The limit of the above as $t\to 0^+$ is $0$ only if $a>5/2$. Next, we show that $a>5/2$ is a sufficient condition. Using the inequality $2uv\leq u^2+v^2$ for $u=|x|^{5/2}$ and $v=|y|$, we have
$$0\leq \frac{|x|^a|y|}{|x|^5+y^2}=|x|^{a-5/2}\frac{|x|^{5/2}|y|}{|x|^5+y^2}\leq \frac{|x|^{a-5/2}}{2}$$
and you can conclude with the squeeze theorem that the limit of the expression is $0$ when $a>5/2$. Hence, the desired values are $a>5/2$.
