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In Thomas's Calculus, 11th Edition, the limit of a function in 2 variables in defined as:

We say that a function $ƒ(x, y)$ approaches the limit $L$ as $(x, y)$ approaches $(x_0, y_0)$ if for every number $\epsilon > 0$, there exists a corresponding number $\delta$ such that for all $(x, y)$ in the domain of $f$, $$ |f(x, y) - L| < \epsilon $$ whenever, $$ 0 < \sqrt{(x - x_0)^2 + (y - y_0)^2} < \delta$$

Can someone help me understand this definition?

I don't understand the significance of "for every number". Is the definition considering large values of $\delta$ and $\epsilon$ as well?

What I understand is for some very small delta there should be a very small value of $\epsilon$.

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    $\begingroup$ It doesn't matter if $\varepsilon$ is very large, but we're primarily interested in what happens when it is small. Your last sentence is completely backward: what we want is that given a small positive number $\varepsilon$ (doesn't matter how small, it can be less than $10^{-1000000000000000000}$) there should be a corresponding $\delta$ such that... $\endgroup$
    – peek-a-boo
    Aug 4, 2019 at 10:16

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Although we usually think of $\epsilon$ and $\delta$ as small numbers, the definition of a limit requires such a $\delta$ to exist for every $\epsilon\gt0$. So, for example we could ask "What $\delta$ causes the function to be at most $1000$ units from the limit $L$?" i.e. $\epsilon=1000$. It may be the case that $\delta$ remains small for large $\epsilon$ but it is also possible that $\delta$ becomes large as it depends on $\epsilon$.

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  • $\begingroup$ So it doesn't matter what the magnitude of $\delta$ and $\epsilon$ is but just that such a $\delta$ should exist for each positive value of $\epsilon$, right? By this definition then, how would we know if the limit doesn't exist? $\endgroup$
    – oshhh
    Aug 4, 2019 at 10:42
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    $\begingroup$ If the limit didn't exist then we would not be able to select any $\delta$ for some $\epsilon\gt0$. $\endgroup$ Aug 4, 2019 at 10:48

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