MHB Problem of the Week #91 - December 23rd, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Use the $\epsilon-\delta$ definition of the limit to prove that $\displaystyle \lim_{(x,y)\to(0,0)} \frac{3x^2y}{x^2+y^2} = 0$.

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Hint: [sp]As you start finding upper bounds for the multivariable inequalities, keep in mind that $x^2\leq x^2+y^2$ since $y^2\geq 0$.[/sp]

 

[Chris L T521]

This week's problem was correctly answered by MarkFL. You can find his solution below.

[sp]We are asked to show that:

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

Switching to polar coordinates (and observing that distance from the origin depends only on $r$), we obtain:

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

Since $$3\cos^2(\theta)\sin(\theta)$$ is bounded for any real $\theta$, we are left with:

$$\lim_{r\to 0} r = 0$$

For any given $\epsilon>0$, we wish to find a $\delta$ so that:

$$|r|<\epsilon$$ whenever $$0<|r|<\delta$$

Thus, to make:

$$|r|<\epsilon$$

we need only make:

$$0<|r|<\epsilon$$

We may then choose:

$$\delta=\epsilon$$

Thus:

$$0<|r|<\delta\implies |r|<\epsilon\implies\lim_{r\to 0} r = 0$$[/sp]

This wasn't the solution I had in mind; I will update this post with my solution later today.