Multivariable limit .... no L'Hopital rule? I am looking a bit at limits for multivariable functions by myself, and I can't figure it out; my book only mentions them shortly, but now I am looking at an "assignments for those interested" and it says 
$$\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy} \cos(x+y)$$
But that's a $0$ over $0$ expression... ? 
Do we have a L'Hopital rule for these types of functions? How would I solve it? 
 A: There is no L'Hospital's rule for multivariable limits. If the limit is to exist, you can let $(x,y)$ approach $(0,0)$ by any path. Hence, if there exists any two paths such that the limits doesn't coincide, then the limit doesn't exists overall. For example, if I make $(x,y)\to (0,0)$ along the line $(t,t)$ (with $t \to 0$), I have: $$\lim_{t \to 0} \frac{\sin(t^2)}{t^2} \cos(2t) = 1 \ \cos(0) = 1.$$
With this, if the limit is to exist, it will be zero. In general, if you use any line $(t, mt)$, the limit will go to $0$. If you use any polynomial path $(t, a_1t + \cdots + a_nt^n)$, the limit will also be zero. Note that no matter how many paths I've checked, I can't conclude that the limit is zero, because it could exist two paths that I haven't checked such that the limits give different values. Usually the strategy is to use epsilons and deltas to prove that it really is zero, or write it as a product of a limited function, by a function which goes to zero... (take the hint!)
A: Hint: $$\lim_{(x,y)\to(0,0)} \frac{\sin (xy)}{xy} = 1$$ 
And $\cos (x + y)$ is continuous. 
A: $$\lim_{t\to0}\frac{\sin t}t=1\qquad\lim_{t\to0}\cos t=1\qquad\lim_{(x,y)\to(0,0)}xy=\lim_{(x,y)\to(0,0)}(x+y)=0$$
