How to find a condition for when a multivariable limit exists If I have the multivariable limit
$$ \lim_{(x,y) \to (0,0)} \frac{x^ay^b}{(x+y)^c} $$
How do I find a general relationship/condition for $a$, $b$, $c$ that results in the limit existing? I've found various specific examples of $a$, $b$, $c$ that make the limit exist, but I don't know what the relationship between them is.
 A: This limit does not exist if $a+b>c$.
First Suppose the limit exists. Then it exists on the path (the line through the origin) $y=m\,x $. And $(x,y) \rightarrow (0,0)$ on this path when $x\rightarrow 0$.
Then
$$\lim_{(x,y)\rightarrow (0,0)}\frac{x^a\,y^b}{(x+y)^c}=\lim_{x\rightarrow 0}\frac{m^b x^{(a+b)}}{(1+m)^cx^c}$$
which blows if $a+b< c$ and depends on the path (on the slope $m$) when $a+b=c$. So, we must have $a+b>c$.
Next we discuss whether the limit exists and equals $0$ when $a+b>c$:
Simply use polar coordinates $x=r\cos{\theta}$, $y= r\sin{\theta}$
and notice that $r\rightarrow 0^{+}$ when $(x,y)\rightarrow (0,0)$.
So, we have
$$\lim_{(x,y)\rightarrow (0,0)}\frac{x^a\,y^b}{(x+y)^c}=\lim_{r\rightarrow 0^{+}}\frac{(\cos{\theta})^{a}(\sin{\theta})^{b} }{(\cos{\theta}+\sin{\theta})^c} r^{(a+b-c)}=0$$
when $a+b>c$. Notice that outside the line $y=-x$ $\cos{\theta}+\sin{\theta}\neq 0 \qquad (1)$.
An edit:
How handle the line $y=-x$ ?
Does the limit not exist because the function is not defined on that line.
The interesting discussion below drew my attention to the following fact: The function $f(x,y):=\frac{x^a\,y^b}{(x+y)^c}$ is not defined on the line $y=-x$. So, the answer above studies the existence of the limit on $\mathbb{R}^{2}\setminus \{(x,y):x+y=0\}$. It does not make sense to study the limit on the path $y=-x$ while the function is not defined thereon. I am now wondering if this problem can be avoided if the question was about the function
$$f(x,y)=\frac{x^a\,y^b}{(x+y)^c},\quad (x,y)\in \mathbb{R}^{2}\setminus \{(x,y):x+y=0\}, \qquad f(x,y)=0\quad \text{otherwise}.$$
An important new edit: As professor Shifrin pointed out. There is an issue here with the use of polar coordinates. One has to be careful with the equality (1).
A: Even if $a+b>c$, when $c>0$ the limit will not exist.
Consider the function along the curve $y=-x+x^\nu$ with $\nu>1$. Then
$$\frac{|x|^a|y|^b}{|x+y|^c} = \frac{|x|^{a+b}|1-x^{\nu-1}|}{|x^{c\nu}|} = |x|^{a+b-c\nu}|1-x^{\nu-1}|.$$
Since we can choose $\nu$ as large as we wish, this quantity will approach $\infty$ as $x\to 0$.
I think it is a mistake to have the kneejerk reaction to switch to polar coordinates. Considering the right polynomial curve is often the right way to establish a limit does not exist.
A: Let me generalise your question, because I think it'll make analysis slightly nicer. Consider the function:
$$g(x, y) = f_1(x, y)^{m_1} \cdot \ldots \cdot f_n(x, y)^{m_n},$$
where $f_i$ takes the form
$$f_i(x, y) = \sin(\theta_i)x + \cos(\theta_i)y.$$
That is, $f_i(x, y) = 0$ is a line through the origin. Moreover, we will assume that $\theta_i \in [0, \pi)$ and are distinct, so as to make each of the lines distinct. To recover your original problem, we let $\theta_1 = \pi/2$, $\theta_2 = 0$, and $\theta_3 = \pi/4$, with $a = m_1$, $b = m_2$, and $c = -m_3$.
Clearly, if $m_1, \ldots, m_n = 0$, then the limit is $1$, provided you don't take $0^0$ to be some value other than $1$ (even if you leave it undefined, most points around $0$ will not lie on any of our $n$ lines, and the function value at these points will be $1$).
If $m_1, \ldots, m_n \ge 0$, with at least one strictly positive, then the factors $f_i^{m_i}$ where $m_i > 0$ will be continuous functions that take the value $0$ at $(0, 0)$, whereas the other functions fall into the previous case. So, the product of the limits will be $0$. This, with the previous case, covers all of the $m_1, \ldots, m_n \ge 0$ cases: the limit does exist.
Let us suppose that $m_1 < 0$. I claim that the limit does not exist, regardless of the other values of $m_i$. Note that, due to the symmetry of the problem, this covers every case of $m_i < 0$.
Let us consider points of the form
$$(x, y) = \left(\frac{\varepsilon}{2} \cos(\theta), \frac{\varepsilon}{2} \sin(\theta)\right),$$
i.e. points whose distance from the origin is $\varepsilon / 2 > 0$. We have,
$$f_i,(x, y) = \frac{\varepsilon}{2}\sin(\theta_i)\cos(\theta) + \frac{\varepsilon}{2}\cos(\theta_i)\sin(\theta) = \frac{\varepsilon}{2}\sin(\theta + \theta_i).$$
Our function $g$ therefore comes to:
$$g(x, y) = \left(\frac{\varepsilon}{2}\right)^{m_1 + \ldots + m_n}\sin^{m_1}(\theta + \theta_1)\ldots \sin^{m_n}(\theta + \theta_n).$$
Consider the function
$$h(\theta) = \sin^{m_2}(\theta + \theta_2)\ldots \sin^{m_n}(\theta + \theta_n).$$
I claim that $h$ is continuous and non-zero at $\theta = -\theta_1$. Note that the factors of $h$ are all continuous, if $\sin(\theta + \theta_i) \neq 0$ for $i > 1$.
If we had $\sin(\theta + \theta_i) = 0$, then $\theta + \theta_i = 0$ or $\theta + \theta_i = -\pi$. These are the only two options for $\theta_i \in [0, \pi)$ and $\theta \in [-\pi, 0)$. If the former were true, then we would have
$$\theta + \theta_1 = 0 = \theta + \theta_i \implies \theta_1 = \theta_i,$$
which contradicts the uniqueness of the $\theta_i$s.
If the latter were true, then $\theta_i = 0$ and $\theta = -\pi$. We would also have $$0 = \theta_1 + \theta = \theta_1 - \pi \implies \theta_1 = \pi,$$
which contradicts $\theta_1 \in [0, \pi)$.
Thus, all the factors of $h(\theta)$ are non-zero and continuous at $-\theta_1$. We can then write:
$$g(x, y) = \left(\frac{\varepsilon}{2}\right)^{m_1 + \ldots + m_n}\sin^{m_1}(\theta + \theta_1)h(\theta),$$
where the first factor is constant, the third factor tends to something non-zero as $\theta \to -\theta_1$, and the second factor blows up due to $m_1 < 0$.
So, we can choose a point of distance $\frac{\varepsilon}{2}$ from the point $\left(\frac{\varepsilon}{2} \cos(\theta_1), \frac{\varepsilon}{2}\sin(\theta_1)\right)$ where $g$ has value that's arbitrarily large (e.g. larger than $100$). This point is at most distance $\varepsilon$ from $(0,0)$, so this point is arbitrarily close to the origin.
Thus, the limit of $g$ as $(x, y) \to (0,0)$ doesn't exist when $m_1 < 0$. Similarly, if any other $m_i < 0$, the limit doesn't exist.
So, in total, the limit exists if and only if $m_1, \ldots, m_n \ge 0$. In your specific case, the answer is, the limit exists if and only if $a \ge 0, b \ge 0, c \le 0$.
