multivariable limit of $\frac{x^2-y^2}{\sqrt{x^2+y^2}}$ Calculate multivariable limit of $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
How to do that? I was trying to figure out any transformations e.g. multiplying by denominator but I do not think it gives me any solution.
 A: Hint. Note, that $\def\abs#1{\left|#1\right|}\abs x \le \sqrt{x^2+y^2}$, hence
$$ \abs{\frac{x}{\sqrt{x^2+ y^2}}} \le 1 
$$
and the same for $y$, that is 
$$ \abs{\frac{x^2-y^2}{\sqrt{x^2 + y^2}}} \le \abs x + \abs y $$
A: $$\lim\limits_{(x,y) \to (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+}\frac{r^2\cos^2\phi-r^2\sin^2\phi}{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}}$$
$$=\lim\limits_{r\to 0^+}\frac{r^2\left(\cos^2\phi-\sin^2\phi\right)}{\sqrt{r^2\left(\cos^2\phi+\sin^2\phi\right)}}$$
$$=\lim\limits_{r\to 0^+}\frac{r^2\left(\cos^2\phi-\sin^2\phi\right)}{\left|r\right|}$$
$$=\lim\limits_{r\to 0^+} r\cos\left(2\phi\right)$$
Now let's attempt to find bounds that are independent of $\phi$
$$0\leq\left|\cos\left(2\phi\right)\right|\leq 1$$
$$0\leq r\left|\cos\left(2\phi\right)\right|\leq r$$
$$ \lim\limits_{r\to 0^+} 0\leq \lim\limits_{r\to 0^+} r\left|\cos\left(2\phi\right)\right|\leq \lim\limits_{r\to 0^+} r$$
$$  0\leq \lim\limits_{r\to 0^+} r\left|\cos\left(2\phi\right)\right|\leq 0$$
Therefore by the squeeze theorem 
$$\lim\limits_{(x,y) \to (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=0$$
A: $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
Transforming $x=r\cos\theta, y=r\sin\theta$.
 $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=\lim_{r \rightarrow 0}\frac{r^2\cos^2\theta-r^2\sin^2\theta}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}}=\lim_{r \rightarrow 0}\frac{r^2(\cos^2\theta-\sin^2\theta)}{r}=\lim_{r \rightarrow 0}{r(\cos^2\theta-\sin^2\theta)}=\lim_{r \rightarrow 0}{0\cdot(\cos^2\theta-\sin^2\theta)}=0.$$
A: $$\bigg|
\frac{x^2-y^2}{\sqrt{x^2+y^2}}\bigg| \leq
\frac{x^2+y^2}{\sqrt{x^2+y^2}} \leq \sqrt{x^2+y^2} $$
