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How would you prove the following limit?

$$\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0$$

I think the best way is using the squeeze theorem but I can't find left expression.

$$0 \le \frac{x^3y}{x^4 + y^2} \le \frac{x^3y}{x^4} \le \frac{x^3y}{x^3} \le y = 0.$$

But I'm not sure I'm right (especially at $\frac{x^3y}{x^4} \le \frac{x^3y}{x^3}).$

If I'm right - I'd glad if you can accept it.

If I'm wrong - can you please correct me?

Thanks in advance!

Billie
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3 Answers3

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Since $|2ab|\le a^2+b^2$, $$ \begin{align} \left|\frac{x^3y}{x^4+y^2}\right| &=|x|\,\left|\frac{x^2y}{x^4+y^2}\right|\\ &\le\frac12|x| \end{align} $$ using $a=x^2$ and $b=y$.

robjohn
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alternately, bracket your limit using its negative and positive absolute value. Since your inequalities show the absolute value going to zero, the lower bracket will go to zero as well.

huh
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Switch to polar coordinates, $x = r \times \cos \theta$, $y = r \times \sin \theta$. Then your limit will be

$r^2 \times \frac{\cos^3 \theta \sin\theta}{r^2\cos^4\theta + \sin^2\theta}$ for $r \to 0$ and an arbitrary $\theta$ path, which is a product of a term going to zero and a finite fraction and hence goes to zero

huh
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  • The path is actually arbitrary, since $\theta$ can be a function of $r$ - no restrictions implied. The idea is that the ratio term is bound (denominator being a sum of positive terms) regardless of the path function $\theta(r)$, so the limit exists and is zero. – huh Feb 12 '14 at 19:10
  • @TooOldForMath I don't see the flaw in the logic either. The points on any path can be expressed as $(x,y) = (r \cos\theta, r \sin\theta)$. If the path approaches to the origin, then the radius $r$ from origin must tend to $0$. As huh says, we are not fixing $\theta$, so we are not just considering lines. I've often used polar coordinates to evaluate multivariable limits, and I certainly hope I haven't been doing it wrong for years! – Viktor Vaughn Feb 12 '14 at 19:11
  • @SpamIam and huh I'm sorry, I made a mistake :-). See my name. – J.R. Feb 12 '14 at 19:16
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    I just thought I saw the usual mistake, which is so often made, where in this case it is completely fine to use polar coordinates, as you have of course explained correctly. – J.R. Feb 12 '14 at 19:18
  • It appears that you are using $r^2$ as the "term going to zero" and $\frac{\cos^3(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$ as the "finite fraction". However, along the path $r=\sin(\theta)$, the "finite fraction" is unbounded. – robjohn May 22 '19 at 13:29