# 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.

• Shifting to polar coordinates does cover all possible paths, because $\theta$ (which doesn't appear in your problem) allows you to find the others paths. But your analysis after the conversion to polar coordinates is incorrect. Apr 15 '15 at 12:12
• You are very correct Timbuc, I missed that one. Thanks. The question still stands though. Apr 15 '15 at 12:13
• And if there is some dependence on the choice of $\theta$ we can infer that the limit does not exist and depends on which path we choose Apr 15 '15 at 12:17
• @OriaGruber: Your approach is correct! Good job. By the way you can use L'hopital's rule to evaluate the limit as $r \to 0$. Apr 15 '15 at 12:24

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.

$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{r \to 0} \frac{r^2}{\sqrt{r^2+1}-1}=\lim_{r \to 0}\frac{r^2(\sqrt{r^2+1}+1)}{r^2+1-1}=\lim_{r \to 0}\sqrt{r^2+1}+1=2$

You were almost there. Use conjugates:

$$\frac{r^2}{\sqrt{r^2+1}-1}=\frac{r^2(\sqrt{r^2+1}+1)}{r^2}=\sqrt{r^2+1}+1\xrightarrow[r\to 0]{}2$$

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 $$\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{x^2+y^2\to0}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{t\to0^+}\frac{t}{\sqrt{t+1}-1}.$$

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.