I'll try to be more elaborate on the point I was trying to get across on the IRC earlier today.
Let's consider the following function, defined in the polar coordinates: $f: [0, \infty) \times (0, 2\pi] \to \mathbb{R}$,
$$
f(r, \theta) = \frac{r}{\theta}
$$
This is well defined, because our angle argument, $\theta$, varies in $(0, 2\pi]$, and not in $[0, 2\pi)$ as usual. Obviously this does not matter at all, because $2\pi$ determines the same angle as $0$. This function determines a function $\mathbb{R}^2 \to \mathbb{R}$ in a natural way, and I will denote that function by $f$ as well, using $x, y$ coordinates instead of $r, \theta$, hoping that will not cause any confusion. Also, $f(0) = 0$ (because the point $(0, 0) \in \mathbb{R}^2$ is represented in polar coordinates as $(0, \theta)$ for any choice of $\theta$, and $f(0, \theta) = \frac{0}{\theta} = 0$).
So, what's the limit:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \;?
$$
Let's try to solve it using your approach:
$$
\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{r \to 0} \frac{r}{\theta} = 0
$$
Easy, wasn't it? And the resulting value does not depend on $\theta$. Let's conclude thus that the limit is $0$ and call it a day.
Right? Wrong! Consider a sequence (in polar coordinates):
$$
x_n = \left(r_n, \theta_n\right) = \left(\frac{1}{n}, \frac{\pi}{n}\right)
$$
Basically, we start with $x_1$ which is $(-1, 0)$ in cartesian coordinates, and we go clockwise, approaching half a turn, and going closer and closer to $(0, 0)$ in process. It's obvious that as a sequence of point in a plane, this sequence tends to $(0, 0)$ (because, well, in cartesian coordinates it's $x_n = (r_n \cos \theta_n, r_n \sin \theta_n) = (\frac{1}{n} \cos(\frac{\pi}{n}), \frac{1}{n} \sin(\frac{\pi}{n}))$, and obviously each component tends to $0$ as $n \to \infty$).
Now, if $\lim_{(x, y) \to (0, 0)} f(x, y)$ was in fact $0$, then we would also have $\lim f(x_n) = 0$, because $x_n \to (0, 0)$. But, surprisingly:
$$
\lim f(x_n) = \lim f(r_n, \theta_n) = \frac{\frac{1}{n}}{\frac{\pi}{n}} = \frac{1}{\pi} \ne 0
$$
What happened? It turns out, that $\lim_{r \to 0} f(r, \theta) = 0$ does not imply that $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$. When we consider the limit as $r \to 0$, we keep $\theta$ fixed, so we only take a limit along a single straight line passing through the origin. This only implies that if we restrict the our function to any such straight line $L$, and consider a function $f|_L: L \to \mathbb{R}$ it will in fact be continuous at $0$. The problem is that the fact that the restricted function $f|_L: L \to \mathbb{R}$ is continuous at origin for every $L$ passing through origin, it's not enough to conclude that our function is in fact continuous at origin.
Although your argument contains a grain of truth, it is not quite correct as it is written, since you wrote that the limit $\lim_{(x,y)\to (0,0)} f(x,y)$, which doesn't exist, is equal to the limit $\lim_{r \to 0} (\cdots)$, which does exist (for $\sin \theta \neq 0$) if you just view it as an ordinary single-variable limit which depends parametrically on a constant $\theta$; in fact, you just computed it yourself like that and wrote that it's equal to $1 + \cot \theta$.
The problem is that you want $\theta$ to be able to vary independently of $r$ as $r \to 0$, so you can't treat $\theta$ as a constant here. It should be viewed as an arbitrary function $\theta(r)$.
In fact, polar coordinates are mainly useful for proving that a limit exist, namely if you can write $f(x,y)$ as a bounded factor times another factor which depends only on $r$ (no $\theta$!) and which tends to zero as $r \to 0$ (really just a single-variable limit here!), then $f(x,y)\to 0$ as $(x,y)\to(0,0)$.
To show that a limit does not exist, you instead find two ways of approaching the point such that you get two different values. In your case, consider $f(t,0)$ and $f(0,t)$ as $t\to 0$, for example.
So actually I don't quite know how I would like to write the argument in a nice way if someone forced me to use polar coordinates in order to show that a limit does not exist! I would probably just write $f(x,y)$ in polar coordinates first,
$$
f(r \cos\theta(r), r \sin\theta(r)) = \cdots,
$$
(no “$\lim$” here) and then say that by making different choices of the function $\theta(r)$ (for example different constant functions!), you can make $f$ approach different values. And I would give examples of two such function $\theta(r)$ which give different limits for $f(r \cos\theta(r), r \sin\theta(r))$ as $r \to 0$.
Best Answer
If you are not comfortable with polar coordinates, your remark
suffices to show that $\left|x \frac{y}{\sqrt{x^2+y^2}} \right|\leq \left| x \right| \leq \sqrt{x^2+y^2}$ and thus $f(x,y) \rightarrow0$ as $(x,y)\rightarrow 0$.