It says in my textbook that $\displaystyle sin(x-y)/cos(x+y) = sin(0)/cos(0) = 0/1 = 0$, so 0 is the limit. For most of the other problems in the section, I have to show that when the limit exists using the definition of limits of two-variable functions (that sqrt(x^2 + y^2) < delta implies that |f(x,y)-L| < epsilon.

Normally I do this with something like: |f(x,y)-L| = .... = .... ≤ |y| or |x| ≤ sqrt(x^2 + y^2) < delta = epsilon.

Or I could show that there is no limit by using a two counterexamples (two functions on the curve sin(x-y)/cos(x+y) that approach different values in f(x,y) near (0,0)).

I found that on the line x=y, lim y->0 $\displaystyle sin(y-y)/cos(y+y) = sin(0)/cos(0) = 0$

But on the curve x = \pi*cos(y), lim y->0 $\displaystyle sin(\pi*cos(y)-y)/cos(\pi*cos(y)+y) = sin(\pi*cos(0)-0)/cos(\pi*cos(0)+0) = sin(\pi)/cos(\pi) = 1/0$, so there is no limit when approaching $\displaystyle sin(x-y)/cos(x+y)$ from the curve $\displaystyle x = \pi*cos(y).$

Doesn't it follow that $\displaystyle sin(x-y)/cos(x+y)$ has no limit at (0,0)? Or do I need to find a counter example of some x=f(y) where lim y->0 $\displaystyle sin(f(x)-y)/cos(f(x)+y)$ actually exists but is not zero?