$$\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^6}$$
I know the limit is 0 if it exists. I want to solve using squeeze theorem but I don't know what to use for the upper function.
How to calculate $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^6}$?
3
$\begingroup$
limits
multivariable-calculus
2 Answers
8
What I'm thinking is that:
$2|xy^3|\le x^2+y^6$ so $\left|\frac{xy^4}{x^2+y^6}\right|\le\frac{|y|}{2}\le\frac{r}{2}\rightarrow 0$, where $r^2=x^2+y^2$
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Please Note
If this is wrong, please let me know why. I don't understand why.
Looking at the denominator let, $u^2=x^2+y^6$. This represents a weird but closed shape enclosing the origin. Parametrize this with $x=u \cos (v)$ and $y=u^{1/3} \sin^{1/3} (v)$. Now approach with $u \to 0^+$ regardless of $v$, closing in on the origin. This gives,
$$=\lim_{u \to 0^+} \frac{u^{7/3} \cos (v) \sin^{4/3} (v)}{u^2}$$
$$=\lim_{u \to 0^+} u^{1/3} \cos (v) \sin^{4/3} (v)$$
$$=0$$
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1To me it looks like what is missing is the proof that $u\rightarrow 0$ when $(x,y)\rightarrow 0$. But this shouldn't be too hard to prove. – 2017-01-20
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0Your proof assumes that if you take $r = \sqrt{x^2 + y^6}$, then we can always find $v$ so that $x = rcos(v)$ and $y = r^{1/3}sin^{1/3}(v)$. Perhaps this is always so, but it requires a proof. – 2017-01-20
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0That part is fine, he just writes the point $(x,y^3)$ in polar coordinates, which can be always done. But it is not proved that $(x,y^3)$ itself converges to $(0,0)$ when $(x,y)$ converges to $(0,0)$ – 2017-01-20
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0Ah, you're right. Sorry for my objections. I didn't realize this problem was that easy to solve. Nice solution and +1. – 2017-01-20