Limit exist for two path test

Question

Why does the limit exist when I apply the two path tests for this function and got two different numbers?

$$\lim_{(x,y)\to(1,-1)}\left(\dfrac{x^2-y^2}{x+y}\right)$$

The two paths I chose were $y = x$ and $y = x - 2$? Thank you!

Answer 1

Your point $(1, -1)$ does not lie on the path $y=x$, so taking a limit along this path makes no sense.

Answer 2

$$\lim_{(x,y)\to(1,-1)}\left(\dfrac{x^2-y^2}{x+y}\right)=$$

$$\lim_{(x,y)\to(1,-1)}\left(\dfrac{(x-y)(x+y)}{x+y}\right)=$$

$$\lim_{(x,y)\to(1,-1)}(x-y)= 2$$

You should get the same result on every path that you choose.