Using polar coordinates to evaluate a multivariable limit

[Jimmy25]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

 

[fzero]

Homework Statement

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)

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.