$\lim_{(x,y) \to (0,0)} f(x,y) = \frac{(e^x-1)y}{x^2+y^2}$
How would you evaluate this limit? I tried many things, I'm pretty sure it's 0 but I'm not sure how to show that.
I tried using squeeze theorem and basically stated:
$y < x^2 + y^2$ Thus, $0 < \frac{|y|}{x^2+y^2} < x^2 + y^2$ and then I multiplied by $e^x - 1$ to get $0 < \frac{(e^x-1)|y|}{x^2+y^2} < (e^x-1)(x^2 + y^2)$ thus when you evaluate all the limits you get 0. I feel like something is wrong here or is this fine? Also, how would I approach this for limits similar to this.
That limit does not exist. In this case you can just use the two path rule: we have that along the path $y = x$ and $y = 2x$ the limits are $\frac{1}{2}$ and $\frac{2}{5}$, respectively, therefore it does not exist. For similar exercises you can try different paths and see if they all converge to the same value (this does not guarantee the existence of the limit, though!): if they do, try to prove that the limit itself converges to that value. There's no one trick to solve every type of exercise like this, you can only really "get it" with practice.