Two variable limit with $\alpha$ and $\beta$. 
$$\lim_{(x,y)\to (0,0)} \frac{(\sin x)^{\alpha} (1-\cos y)^{\beta}}{(y^2+x^2)^4}$$
$\alpha ,\beta \ge 0$, for what values of $\alpha , \beta$ the limit exists?

My work: 
First of all, just trying to get an idea, I know that $\sin(x) \sim x$ and $1-\cos y \sim \frac{y^2}{2}$, near $(0,0)$, so I would say that this limit is $\frac{x^{\alpha}\frac{y^{2\beta}}{2^{\beta}}}{(x^2+y^2)^4}$, and since I have a homogeneous polynomial in the denominator with a power of $8$, I can say that $\alpha +2 \beta > 8$ are the values that this limit exists for them. 
Now this was just my own thinking and nothing formal, and I'm unsure if it's right or not, like am I allowed to jump to the next limit that I showed? did I get a correct answer or did I just think nonsense? 
I would appreciate any feedback and how to do this formally and not just by what I did (incase if it's right too).
 A: Yes, you're right, but if you want to be precise you should keep track of how big the remainder terms are when making these approximations. Let's call the numerator $f(x,y)$. Then, as $(x,y)\to (0,0)$,
\begin{align}
f(x,y)&=\left[x+ O(x^3)\right]^{\alpha}\left[\frac{y^2}{2}+O(y^4)\right]^{\beta}\\
&=\frac{x^{\alpha}y^{2\beta}}{2^{\beta}}\underbrace{[1+O(x^2)]^{\alpha}[1+O(y^2)]^{\beta}}_{:=\rho(x,y)}
\end{align}
Therefore,
\begin{align}
\frac{f(x,y)}{(x^2+y^2)^{4}}&=\frac{x^{\alpha}y^{2\beta}}{2^{\beta}(x^2+y^2)^4}\cdot \rho(x,y),
\end{align}
whereby from the definition of $\rho$, it is evident that $\lim\limits_{(x,y)\to(0,0)}\rho(x,y)=1$. Now, as you mention,

*

*if $\alpha+2\beta>8$, the first factor approaches $0$; so the fact that $\rho(x,y)\to 1$ (in particular it is bounded near the origin) implies by the squeeze theorem that the product also approaches $0$.

*If $\alpha+2\beta=8$, then the limit does not exist because the limit along the line $x=y$ is non-zero... it is infact equal to $\frac{1}{2^{\beta+4}}$. On the other hand, $\alpha+2\beta=8>0$ and $\alpha,\beta\geq 0$ imply that one of these quantities is strictly positive. If $\alpha>0$, then take the limit along the line $x=0$, while if $\beta>0$ take the limit along $y=0$. In any case, we have two paths, along one we have a non-zero limit, along the other we have a zero limit.

*If $\alpha+2\beta<8$, then the absolute value of the product approaches $\infty$.


Just in case you're not familiar with the big O notation, when I write $O(x^2)$ above, what I mean is that it is equal to a function $\phi(x)$, such that there exist $M>0,\delta>0$ such that for all $|x|\leq \delta$, we have $|\phi(x)|\leq M|x^2|$.
Note that nowhere in my answer did I use any $\sim$ symbols. They are all truly equal signs (once we understand the meaning of big O), and each equal sign captures exactly "how large" the remaining terms are. So, this really is completely formal. If this is your first time seeing this, I would suggest reading section 3.5, on infinitesimals, of Loomis and Sternberg's Advanced Calculus (atleast up to theorem 5.1).
