Limits — Question 6

Question 6

Let $$D=\{(x,y):x\ge 0,\ 0\le y\le\sqrt 3\,x\},\qquad f(x,y)=\frac{x^2-y^2}{x^2+y^2}.$$ Find the limit of $f$ as $(x,y)\to(0,0)$ within $D$, or prove it does not exist.

Tasks

1. Describe the allowed directions.

2. Compare two valid boundary paths.

3. State the relative-limit conclusion.

Question 6 – Solution

Strategy. A relative limit uses only paths in the domain, but all such paths must still agree.

Step 1: Directions The wedge is $0\le\theta\le\pi/3$. In polar form, $$f=\frac{r^2\cos^2\theta-r^2\sin^2\theta}{r^2}=\cos 2\theta.$$

Step 2: Boundary paths Along $y=0$ ($\theta=0$), $f=1$. Along $y=\sqrt 3x$ ($\theta=\pi/3$), $$f=\cos(2\pi/3)=-\frac 12.$$ Both paths stay in $D$ and approach the origin.

Conclusion. Because the two permitted approaches disagree, $$\boxed{\displaystyle \lim_{\substack{(x,y)\to(0,0)\\(x,y)\in D}}f(x,y)\text{ does not exist}.}$$