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How can I prove that

$\lim_{(x,y)\to (0,0)} \frac{\left | x \right |^{\frac{3}{2}}y^{2}}{x^{4} + y^{2}} \rightarrow 0\;?$

Thanks!

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    @Hila: That's no excuse for you to do it. ;-) Anyway, writing "$f(x) \to A$ as $x \to a$" is not looser than writing "$\lim_{x\to a} f(x)=A$", but please don't mix the two.2011-07-11

2 Answers 2

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For $y \neq 0$, $ 0 \le \frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} = \frac{{|x|^{3/2} }}{{x^4 /y^2 + 1}} \le \frac{{|x|^{3/2} }}{1} = |x|^{3/2} \to 0. $

EDIT: In retrospect, simply note that $0 \le \frac{{y^2 }}{{x^4 + y^2 }} \le 1$, for $(x,y) \neq (0,0)$, to conclude that $\frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} \to 0$ as $x \to 0$, independently of $y$.

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    Excellent totally forgot you can raise both sides of the final inequality by $3/2$. I think in the post above I meant take $\epsilon = \text{min}\{1, \delta\}$.2011-07-07
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I assume you want to show that

$\lim_{(x,y)\rightarrow (0,0)}\frac{|x|^\frac{3}{2}y^2}{x^4+y^2}= 0.$ Although sometimes it is not the nicest way, switching to polar coordinates is a general approach that will solve these kinds of limits. (and we don't have to think too much!) Let $x=r\cos \theta$, $y=r\sin\theta$. Then your limit is $\lim_{(x,y)\rightarrow(0,0)}\frac{|x|^{\frac{3}{2}}y^{2}}{x^{4}+y^{2}}=\lim_{r\rightarrow0}\frac{r^{\frac{3}{2}}|\cos\theta|^{\frac{3}{2}}r^{2}\sin^{2}\theta}{r^{2}\left(r^{2}\cos^{4}\theta+\sin^{2}\theta\right)}=\lim_{r\rightarrow0}r^{\frac{3}{2}}\frac{|\cos\theta|^{\frac{3}{2}}}{\left(r^{2}\cos^{4}\theta/\sin^{2}\theta+1\right)}.$

Now, since

$0\leq \frac{|\cos\theta|^{\frac{3}{2}}}{\left(r^{2}\cos^{4}\theta/\sin^{2}\theta+1\right)}\leq |\cos\theta|^\frac{3}{2}\leq 1$

we can apply the squeeze theorem, and since $\lim_{r\rightarrow 0}r^\frac{3}{2}=0$, we see that the original limit is $0$.

Hope that helps,

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    @Eric Naslund: See http://www.amazon.com/Counterexamples-Analysis-Dover-Books-Mathematics/dp/04864287532011-07-06