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How can I find the following limit using the Squeeze theorem?

$$\lim_{x,y\to 0} \frac{5xy^2}{x^2+y^2}$$

I am not sure what to use for the lower and upper bounds.

PabloDPO
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3 Answers3

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$$\Bigg|\frac{5xy^2}{x^2+y^2} \Bigg|\le\Bigg|\frac{5xy^2}{y^2} \Bigg|=|5x|$$

luka5z
  • 6,359
  • That would be the upper bound, but what about the lower bound? Or can I use 0 for the lower bound? – PabloDPO Mar 12 '17 at 10:01
  • $|f(x,y)|\le a \iff f(x,y)\le a \wedge f(x,y)\ge -a$... So your upper bound is $|5x|$ and lower is $-|5x|$. Both go to $0$. – luka5z Mar 12 '17 at 10:08
4

Note that $${ x }^{ 2 }-2xy+{ y }^{ 2 }\ge 0\\ { x }^{ 2 }+{ y }^{ 2 }\ge 2xy\\ \frac { 1 }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } \le \frac { 1 }{ \left| 2xy \right| } $$ so $$ \\ \\ \\ \\ \left| \frac { 5xy^{ 2 } }{ x^{ 2 }+y^{ 2 } } \right| \le \left| \frac { 5x{ y }^{ 2 } }{ 2xy } \right| =\frac { 5 }{ 2 } \left| y \right| $$

haqnatural
  • 21,578
2

You could try converting into polar co-ordinates which gives

$$\lim_{r \to 0}\frac{5r^3\cos(\theta)\sin^2{\theta}}{r^2(\cos^2{(\theta)}+\sin^2(\theta))}=\lim_{r \to 0}5r\cos(\theta)\sin^2{\theta}=0$$

$$|\sin(\theta)| \leq1$$ $$|\cos(\theta)| \leq1$$ $$|\sin^2(\theta)\cos(\theta)| \leq1 $$

Then $$|5r\sin^2(\theta)\cos(\theta)| \leq 5r$$

Since $\lim_{r \to 0}(-5r)=\lim_{r \to 0}(5r)=0 $

Thus $r$ is bounded which means our step was justified even though $\theta $

is arbitrary

By Squeeze theorem

$$\lim_{r \to 0}5r\cos(\theta)\sin^2{\theta}=0$$

EDIT : AS mentioned by Yves Daoust, $$|\sin^2(\theta)\cos(\theta)|\leq \frac{2\sqrt3}{9}$$ which can be verified by taking derivatives and also givs a tighter bound.

LM2357
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