# Limits with two variables, finding k such that limit exists

Find the biggest number $k$, such that the limit $$\lim_{(x,y)\to(0.0)} \frac{x^{15}y^{23}}{(x^2 +y^2)^p}$$

exists for all $p < k$

I was thinking that if we're left with x's and/or y's in the numerator and denominator, we have an expression in an undetermined form, $0 \over 0$ so i though that $k = 7.5$ would be correct, seeing as that would give a $0\over 0$ expression, atleast from what I've calculated.

Write $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then \begin{gather*} \lim_{(x,y)\to(0,0)} \frac{x^{15}y^{23}}{(x^2+y^2)^p} = \lim_{r \to 0} \frac{r^{38} \cos^{15}\theta \sin^{23}\theta}{r^{2p}} \end{gather*} In order to have a limit, you need $38 > 2p$.
• @Pame No. First, even fixing $\theta$, the limit would be infinity which should be considered as "limit does not exist". Otherwise, if you allow the limit to be infinity, you can still choose $\theta = 0$ to "make" the limit to be $0$. Namely different subsequences give you different limits, the limit does not exist. Commented Feb 5, 2018 at 19:41
let $u = \max(|x|,|y|)$
$|\frac {x^{15}y^{23}}{(x^2+y^2)^p}| \le\frac {u^{38}}{u^{2p}}$
use that $$\frac{x^2+y^2}{2}\geq |xy|$$ so $$\frac{1}{x^2+y^2}\le \frac{1}{2|xy|}$$