tough GRE subject problem what is the smallest value for $n$ for which the following limit exists for all $r \geq n$?
$$\lim_{(x,y) \rightarrow (0,0)}\frac{x^r}{\vert x \vert^2+\vert y \vert^2}$$
I thought of using LHopitals rule but get stuck differentiating the absolute values. I wonder if there's a trick I am unaware of. So the comments are saying to rewrite as
$$\lim_{(x,y) \rightarrow (0,0)}\frac{x^r}{x^2+ y^2}$$
Then by LHopitals Rule (twice) I get
$$\frac{r(r-1)x^{r-2}}{4}$$
But I get stuck here.
 A: You cannot use anything like L'Hopital since this does not apply to multivariate limits. The trick, as suggested in the comments, is to switch to polar coordinates: $x=u\cos(v)$ and $y=u\sin(v)$ so that $(x,y) \to (0,0)$ is equivalent to $u \to 0$ (and $v$ left arbitrary). Then $|x|^2+|y|^2=x^2+y^2=u^2$. So our limit is now:
$$ \lim\limits_{(u,v) \to (0,v)} \dfrac{u^r\cos^r(v)}{u^2} = \lim\limits_{(u,v) \to (0,v)} u^{r-2}\cos^r(v)$$
If $r>2$, then $u^{r-2}\cos^r(v)$ is bounded by $\pm u^{r-2}$ and  since $\pm u^{r-2} \to 0$ as $u \to 0$, we have that $u^{r-2}\cos^r(v) \to 0$ as $(u,v)$ heads to the polar origin.
If $r<2$, approaching along the ray $v=0$ (i.e., the positive $x$-axis), we are left with $u^{r-2}$ which blows us. Approaching along other rays can yield other values as well. So the limit does not exist for $r<2$.
Finally at $r=2$, we have $\cos^r(v)$. Since this depends on the choice of $v$, the limit does not exist. Or you could go back to the original expression: $\dfrac{x^2}{x^2+y^2}$ and see that along the $x$-axis (i.e., $y=0$) this is 1 and along the $y$-axis (i.e., $x=0$) this is $0/(0+y^2)=0$.
In summary, the limit exists (and equals 0) exactly when $r>2$.
A: For $r >2$, $r \in \mathbb{N}, $ $(x,y) \not=(0,0)$.
$0 \le |\dfrac{x^r}{x^2+y^2}|\le |\dfrac{x^r}{x^2}| = |x|^{r-2}$.
$r=0,1,2$ are ruled out (why?).
