Is this a valid proof? Find $\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$ I "solved" this limit using polar coordinates, but my question is - is this a definite proof that the limit exists? Or maybe there is some path that I missed when I transformed to polar coordinates? 
$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{r \to 0} \frac{r^2}{\sqrt{r^2+1}-1}=\lim_{r \to 0}\frac{r^2}{r\sqrt{1+\frac{1}{r^2}}-1}=\lim_{r \to 0}\frac{r}{\sqrt{1+\frac{1}{r^2}}-1}=0$
So via polar coordinates, the limit is zero. But maybe there is a path I missed and the limit via that path does not tend to zero?
Please note I am not asking just on this problem. My question is a general question - does polar coordinates shift cover all possible paths?
Edit - Please Read: I realize I made a mistake, The $r$ in the denominator can't be cancelled it, I was careless and missed it. Thanks for the input. I am still very much interested in knowing if polar coordinates cover all paths, which is the original point of this question. Not to solve this particular problem.
 A: I think you made a mistake for the last part. And if you apply polar coordinate properly ( in general it may not simplify the problem though), it will give you the correct answer.
$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{r \to 0} \frac{r^2}{\sqrt{r^2+1}-1}=\lim_{r \to 0}\frac{r^2(\sqrt{r^2+1}+1)}{r^2+1-1}=\lim_{r \to 0}\sqrt{r^2+1}+1=2$
A: You were almost there. Use conjugates:
$$\frac{r^2}{\sqrt{r^2+1}-1}=\frac{r^2(\sqrt{r^2+1}+1)}{r^2}=\sqrt{r^2+1}+1\xrightarrow[r\to 0]{}2$$
A: Yes, polar coordinates can do, as any point $(x,y)$ can be expressed as some $(r\cos\theta,r\sin\theta)$.
For example, 
$$\frac{x-y}{x+y}=\frac{r\cos\theta-r\sin\theta}{r\cos\theta+r\sin\theta}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}$$ has no limit as the last expression is undeterminate.
Your given example is a little different in that $\theta$ does not appear explicitly, hence it is actually a $1D$ limit
$$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{x^2+y^2\to0}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{t\to0^+}\frac{t}{\sqrt{t+1}-1}.$$
A: Maybe you made a mistake for the last part, when you semplify $r$. I invite you to consider this other approach. We have that
$$\sqrt{1+x}-1\sim \frac{x}{2} \ \ (x\rightarrow 0)$$
Thus
$$\sqrt{1+x^2+y^2}-1\sim \frac{x^2+y^2}{2} \ \ (x,y\rightarrow 0)$$
Then the limit is 2. We can verify it.
$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|=\left|\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{x^2+y^2}-2\right|$$
that is
$$\left|\sqrt{x^2+y^2+1}-1\right|\leq \left|\frac{x^2+y^2}{2}+1-1\right|\leq \frac{x^2+y^2}{2}$$
and it tends to zero.
