Why can't we solve $\lim\limits_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}$ through dividing by xy in both numerator and denominator? When solving this limit, we use methods like $\epsilon$ - $\delta$ definition and polar coordinate convertion. And I wonder if we can solve it by dividing by $xy$ in both numerator and denominator like this 
$\lim\limits_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} = \lim\limits_{(x, y) \to (0, 0)} \frac{1}{\sqrt{\frac{1}{y^2} + \frac{1}{x^2}}} = 0$ 
Thanks in advance!
 A: No, not really. The reason is the usual one that you can't divide by zero, so you cannot perform that division when $xy=0$. But that is on either coordinate axes. In particular it includes lots of points outside of just $(0,0)$.
You can see that the first expression is only undefined when both $x$ and $y$ are equal to zero, i.e. only at the point $(0,0)$ whereas the expression that you have written is undefined when either $x$ or $y$ is equal to zero.
A: You can actually divide by $x$ and $y$, with a grain of salt:$\def\sgn{\operatorname{sgn}}$
$$\begin{align}
f(x,y) = \frac{xy}{\sqrt{x^2 + y^2}} 
&= \frac{xy}{|x||y|\sqrt{\frac1{x^2} + \frac1{y^2}}} \\
&= \frac{\sgn(x)\sgn(y)}{\sqrt{\frac{1}{y^2} + \frac{1}{x^2}}} \tag 1
\end{align}$$
which is due to $\sqrt {x^2}=|x|$.  The function $\sgn x$ denotes the sign of $x$:
$$\sgn(x) = \begin{cases}
+1,& x > 0 \\
-1,& x < 0 \\
0,& x = 0 \\
\end{cases}$$
Now the denominator of $(1)$ goes to $\infty$ as $(x,y)\to(0,0)$ and the numerator is bounded irrespective of $x$ and $y$, thus $f(x,y)\to0$.
However there is a loophole:  $f(x,y)$ is defined for any $(x,y)\neq(0,0)$, for example $f(0,y)$ is perfectly fine provided $y\neq0$.  But when you factor out $|x|$ and $|y|$ in the first step, this requires that neither $x$ nor $y$ may be 0.
A way around that loophole could be to use polar coordinates $(x,y) = r(\cos \phi,\sin\phi)$ and using $\sin^2\phi + \cos^2\phi = 1$:
$$\begin{align}
\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2 + y^2}} 
&= \lim_{r\to0^+} \frac{r\cos\phi\cdot r\sin\phi}{\sqrt{r^2\cos^2\phi +r^2\sin^2\phi}} \\
&= \lim_{r\to0^+} \frac{r^2 \cos\phi \sin\phi}{\sqrt{r^2}} \\
&= \lim_{r\to0^+} r \cos\phi \sin\phi \\
&= 0\\
\end{align}$$
A: Others already answered your question so I'm just going to present another approach. You can notice that
$$\left|\frac{xy}{\sqrt{x^2+y^2}} \right| = |x|\cdot \frac{|y|} {\sqrt{x^2+y^2}} \leq |x| \longrightarrow 0$$
as $(x,y)\to (0,0)$.
