# Homework Help: Using polar coordinates to evaluate a multivariable limit

1. Sep 28, 2010

### Jimmy25

1. The problem statement, all variables and given/known data

When you substitute polar coordinates into a multivariable limit, do you treat theda as a constant when evaluating? (I know how to use polar coordinates to evaluate a limit but haven't learned what they are yet)

2. Relevant equations

3. The attempt at a solution

2. Sep 29, 2010

### fzero

Generally you wouldn't treat $$\theta$$ as a constant because the most important concept in taking a multivariable limit is that the limit doesn't exist if it depends on the path you take to the point.

As an example, consider

$$\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y}{x^4+y^2} = \lim_{r->0} \frac{r\cos^2\theta\sin\theta}{r^2 \cos^4\theta + \sin^2\theta} = \lim_{r->0} \frac{r \cos^2\theta}{\sin\theta}.$$

If we just take $$r\rightarrow 0$$, we find that this vanishes. However, if we also take $$\theta\rightarrow 0$$ at the same rate as r, we find $$\cos^2(0)=1$$. Therefore the limit does not exist.