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How can I prove that

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


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Do you mean $(x,y) \rightarrow 0$? Are $x$ and $y$ real? What did you try? – Listing Jul 6 '11 at 16:49
You're mixing notations! One says that the limit is (or equals) zero, or that the function tends to zero. But not that "the limit tends to zero". – Hans Lundmark Jul 7 '11 at 21:53
@Hans Thanks! Our professor uses really loose notation. – Hila Jul 11 '11 at 9:39
@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. – Hans Lundmark Jul 11 '11 at 14:20

2 Answers 2

up vote 7 down vote accepted

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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Of course, this is in case $x \to 0$... – Shai Covo Jul 6 '11 at 16:57
And this is quite obviously the case, in view of the absolute value sign around $x$ and the fact that the limit is $\infty$ when $x,y \to \infty$ with $y=x^2$. – Shai Covo Jul 6 '11 at 18:35
say you wanted to use $\epsilon - delta$, I think you would need to say something like "if we agree that our points $x,y$ are within a ball of radius 1 about the origin", then $|x|^ {3/2} < \sqrt{x^2 +y^2}$. So if $\sqrt{x^2 + y^2}< \delta$, then from your manipulations and taking $\epsilon = \text{min}\{1,\sqrt{x^2 + y^2}\}$ should complete the proof. – user38268 Jul 6 '11 at 23:42
@D Lim: Given $\varepsilon > 0$, you can simply take $\delta = \varepsilon ^ {2/3}$, so if $0 < \sqrt{x^2+y^2} < \delta$, then $$ \bigg|\frac{{|x|^{3/2} y^2 }}{{x^4 + y^2 }} - 0\bigg| \le |x|^{3/2} \le (\sqrt {x^2 + y^2 } )^{3/2} < \delta ^{3/2} = \varepsilon . $$ – Shai Covo Jul 7 '11 at 0:21
Excellent totally forgot you can raise both sides of the final inequality by $3/2$. I think in the post above I meant take $\epsilon = \text{min}\{1, \delta\}$. – user38268 Jul 7 '11 at 0:52

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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Actually, one needs to be careful about using polar coordinates to compute the limits. While this method works most of the time, it does not always work. As an example, the limit of $\frac{x^2y}{x^4+y^2}$ as $(x,y)$ goes to $(0,0)$ does not exist (let $(x,y)$ approach the origin along the line $x = 0$ and along the parabola $y=x^2$. But if you were to use polar coordinates to compute this limit, you'd conclude that the limit is $0$ which would be incorrect. Basically, using polar coordinates allows you to check what happens when $(x,y)$ goes to origin along all lines passing through the origin – algebra_fan Jul 6 '11 at 19:15
but it does not tell you how the function behaves when $(x,y)$ approaches the origin along more complicated curves (e.g. it does not tell you how the function $\frac{x^2 y}{x^4+y^2}$ behaves when $(x,y)$ goes to $(0,0)$ along the parabola $y=x^2$. – algebra_fan Jul 6 '11 at 19:17
@Algebra_fan: Yes it does always work, when applied correctly. I don't understand how you would conclude that limit of your above function is zero. Polar coordinates are usually the best way to understand more complicated curves such as the one you mention. (Or what about limits that converge on every quadratic curve but diverge on cubics? etc..) For your example: Suppose $\sin t\neq 0$. Then $$\frac{x^{2}y}{x^{4}+y^{2}}=\frac{r^{3}\cos^{2}t\sin t}{r^{4}\cos^{4}{t}+r^{2}\sin^{2}t}.$$ – Eric Naslund Jul 6 '11 at 20:28
(continued)$$=r\frac{\cos^{2}t\sin t}{r^{2}\cos^{4}t+\sin^{2}t}=\frac{r}{\sin t}\frac{\cos^{2} t}{r^{2}\frac{\cos^{4} t}{\sin^{2} t}+1}.$$ Because the rightmost fraction is both bounded above and below when $r$ and $t$ are small, this tells us that if $\sin t$ goes to zero faster then $r$, the limit will explode, and hence does not exist. – Eric Naslund Jul 6 '11 at 20:29
Thank you for your response. When I wrote my previous comment, I (incorrectly)assumed that $t$ is fixed, i.e. for every fixed $t$, the limit $\frac{r^3\cos^2t\sin t}{r^4\cos^4 t+r^2 \sin^2t}$ is equal to $0$. But as you showed, we need to let both $r$ and $t$ go to $0$ in order to see that the limit does not exist. So my previous comment was not quite correct. Once again, thank you. – algebra_fan Jul 6 '11 at 21:26

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