Limit of given fraction approaching to zero

Question

Show that $\lim_{(x,y) \to (0,0)} f(x,y) = \frac{x^3(y^3 + \pi)}{x^2+y^2}$ is equal to zero.

My progress: I have two ideas related to this problem. Initially, I thought about using polar coordinates, but quite uncertain whether it is right approach for this problem or not. Alternatively, I am thinking about squeeze theorem: by AM-GM and triangle inequality, I managed to find a function larger than given $f$ with variables $x,y$ only. Does that imply limit of $f$ will be zero?

Does this answer your question? Multivariable limit proof: $\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$ — SescoMath, May 05 '20 at 03:22

user159888 · Accepted Answer · 2020-05-05T04:09:52.810

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Polar form: $\lim_{(x,y) \to (0,0)} \frac{x^3(y^3 + \pi)}{x^2+y^2}=\lim_{r\rightarrow 0} r\cos^3\theta( r^3\sin^3\theta+\pi)=0. $

edited May 05 '20 at 04:09

answered May 05 '20 at 03:37

user159888

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score 3 · Answer 2 · answered May 05 '20 at 03:20

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Sqeeze it: $0<\left|\frac{x^3(y^3 + \pi)}{x^2+y^2}\right|< x^2y^2+x\pi.$ Take the limits now.

answered May 05 '20 at 03:20

user159888

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How about polar coordinates approach? Is that applicable in this problem? – BlizzardWalker May 05 '20 at 03:31
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Yes. That is also very direct. – user159888 May 05 '20 at 03:33
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I did the polar method. Kindle see. – user159888 May 05 '20 at 03:40

Limit of given fraction approaching to zero

2 Answers2