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

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

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

-----

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

#### Chris L T521

##### Well-known member
Staff member
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.

Status
Not open for further replies.