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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.

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3 Answers 3

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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 $38 > 2p$.

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  • $\begingroup$ Isn't it fine to have the denominator go towards 0? Since it's never actually 0 it just approaches 0. $\endgroup$
    – Pame
    Commented Feb 5, 2018 at 19:36
  • $\begingroup$ @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. $\endgroup$ Commented Feb 5, 2018 at 19:41
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let $u = \max(|x|,|y|)$

$|\frac {x^{15}y^{23}}{(x^2+y^2)^p}| \le\frac {u^{38}}{u^{2p}}$

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use that $$\frac{x^2+y^2}{2}\geq |xy|$$ so $$\frac{1}{x^2+y^2}\le \frac{1}{2|xy|}$$

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