# Epsilon-Delta definition of double limit.

Let $$f: \mathbb{R}^2 \rightarrow M$$, with $$(M,d)$$ some metric space.

My question is the following: how would one translate the limit $$\lim_{x \rightarrow a} \lim_{y \rightarrow b} f(x,y) = l$$ in $$\epsilon - \delta$$ notation?

My guess would have been $$\forall \epsilon > 0 \ \exists \delta_1, \delta_2 > 0 \ \operatorname{s.t.} \vert x - a \vert < \delta_1 \ \land \ \vert y - b \vert < \delta_2 \ \Rightarrow \ d(f(x,y),l) < \epsilon.$$ However, I am told that this formulation contains a hidden assumption of "uniformity of the first limit in x". Could somebody pleas reply with the correct $$\epsilon - \delta$$ formulation, and possibly elaborate on this hidden "uniformity" assumption?

Please notice that I am not interested in $$\lim_{(x,y) \rightarrow (a,b)} f(x,y) = l,$$ but rather in the formulation above, where the two variables are treated separately, i.e. they tend to $$a$$ or $$b$$ independently in the norm of $$\mathbb{R}$$, and not jointly to $$(a,b)$$ in the (usual) norm of $$\mathbb{R}^2$$.

$$\forall \epsilon > 0: \exists \delta > 0: \forall x: (0 < |x-a|<\delta \implies |l - \lim_{y \to b} f(x,y)| < \epsilon)$$
provided that $$\lim_{y \to b} f(x,y)$$ exists for $$x$$ in a neighborhood of $$a$$.