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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 $.

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This may be unsatisfying but I think it is the best you can get:

$$\forall \epsilon > 0: \exists \delta > 0: \forall x: (0 < |x-a|<\delta \implies |l - \lim_{y \to b} f(x,y)| < \epsilon)$$

provided that $\lim_{y \to b} f(x,y)$ exists for $x$ in a neighborhood of $a$.

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