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

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Chris L T521

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Jan 26, 2012
995
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:
As you start finding upper bounds for the multivariable inequalities, keep in mind that $x^2\leq x^2+y^2$ since $y^2\geq 0$.


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL. You can find his solution below.

We are asked to show that:

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

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

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

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

\(\displaystyle \lim_{r\to 0} r = 0\)

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

\(\displaystyle |r|<\epsilon\) whenever \(\displaystyle 0<|r|<\delta\)

Thus, to make:

\(\displaystyle |r|<\epsilon\)

we need only make:

\(\displaystyle 0<|r|<\epsilon\)

We may then choose:

\(\displaystyle \delta=\epsilon\)

Thus:

\(\displaystyle 0<|r|<\delta\implies |r|<\epsilon\implies\lim_{r\to 0} r = 0\)


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