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I've perused similar questions, but it seems to me like epsilon-delta proofs can vary greatly from equation to equation. After attempting several different problems, I'm still not fully understanding how an "epsilon-delta" proof works.

Problem: Let $f(x, y) = x^2 + xy + y$. Give an "epsilon-delta" proof that $lim_{(x,y) -> (1,1)} f(x,y) = 3$.

The given solution is that 𝛿 = min(1, ${\frac{ε}{7}}$).

I understand that I'm looking to prove that, for $lim_{(x -> a)} f(x) = b$, if 0<||x-a|| < 𝛿 then || f(x) - b|| < ε, for some ε > 0 and 𝛿 = 𝛿(ε), and that I first need to find 𝛿 in relation to ε.

Attempt at a solution:

Let |y| ≤ |x|

Then, $|| y^2 + y^2 + y - 3|| ≤ ||x^2 + xy + y - 3)|| < ε$

$|| y^2 + y^2 + y - 3|| = || 2(y-1)^2 + 4(y-1)+(y-1)||$

and $0 ≤ ||(y-1)||< ||(x, y) - (1, 1)|| < 𝛿$

If I set δ = ||(y-1)||, I can get: 2δ + 4δ + δ < ε

so δ < $\frac{ε}{7}$, but I'm not sure where the number 1 in the solution comes from.

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You have:

$|f(x,y) - 3|\le|2(y-1)^2 + 4(y-1) + (y-1)|$

Then you say when $|y-1|\le\|(x,y)-(1,1)\|<\delta \implies |2(y-1)^2 + 4(y-1) + (y-1)| <2\delta + 4\delta + \delta$

But that is only true if $\delta \le 1$

i.e. otherwise you would have a $\delta^2$ term.

Nothing keeps you from capping $\delta$ at $1,$ so go ahead and do it.

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