Calculus – Understanding Epsilon-Delta Confusion

calculus

I don't understand the epsilon delta definition of a limit

"According to the formal definition above, a limit statement is correct if and only if confining x to d units of c will inevitably confine f(x) to epsilon units of L."
So, if we can confine x to infinitely small delta units of c, we can confine f(x) to infitely small epsilon units of L. Like, constricting f(x) to L ?

Is that a right way of explaining what the general idea of the epsilon-delta method is ?

Best Answer

The definition I learned is

$ lim_{x \rightarrow y}{f(x)} = c $ if and only if for every $ \epsilon > 0 $, there is a $ \delta > 0 $ such that $ 0 < |x - y| < \delta$ implies $ | f(x) - c | < \epsilon $ for all $x$

This is still a little bit technical, lets see what it means. I think it's easiest to read $ 0 < |x - y| < \delta $ as "the distance between $x$ and $y$ is between $0$ and $\delta$".

If we use this interpretation, the definition becomes: $lim_{x \rightarrow y}{f(x)} = c$ if and only if for every $ \epsilon > 0 $, there is a $\delta > 0$ such that if the distance between $x$ and $y$ is between $0$ and $\delta$, the distance between $f(x)$ and $c$ is less than $\epsilon$.

This basically means that $f(x)$ gets arbitrarily close to $c$ (and I think this expression is still used sometimes as a more informal definition) without necessarily becoming $c$ (as James S. Cook pointed out). Suppose $f(x)$ does not get arbitrarily close to $c$, i.e., there is some constant $d$ such that $f(x)$ will stay away from $c$ with at least distance $d$. Then we can show the limit definition does not hold: take $\epsilon = d$. Now there doesn't exist a $\delta$ such that $0 < |x - y| < \delta \rightarrow |f(x) - c| < \epsilon $ (because we just assumed that $f(x)$ would never get closer to $c$ than distance $d$, and remember that this happening is equivalent to $|f(x) - c| < \epsilon = d$).

What may be helpful too is a (trivial) proof using the limit definition. Usually in these proofs, you take an $epsilon$, and return a $\delta$ for which you have proved that $ 0 < |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ (when dealing with continuous functions you usually just use $|x - y| < \delta$ in your proof). I prove that $\lim_{x \rightarrow y} x = y$ (or, equivalently $f(x) = x$ and $c = y$ in the original formulation).

Take $\delta = \epsilon$. Then $ |x - y| < \delta \rightarrow | f(x) - y | = | x - y | < \delta = \epsilon $. So the condition that there exists some $\delta$ for which $ |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ holds is true, as I've just shown. More advanced proof usually use a similar logic, but the expression for $\delta$ and working out the $| f(x) - c | < \epsilon$ can become quite hard. This is why other theorems are often used (for continuous functions, $ lim_{x \rightarrow y} f(x) = f(y) $, and compositions, products, sums of continuous functions are again continuous, which can help you out very often).

Also, the definition requires that something must hold for every $\epsilon$. Sometimes, teachers explain this as a game: you can choose $\epsilon$ freely, and you can give a procedure to show that something (there exists a $\delta$ such that...) will hold, then we can say $lim_{x \rightarrow y} f(x) = c$. So, if you win this game by giving such a procedure, you basically got the recipe for a proof!

If you cannot win, it suffices to give a single $\epsilon$ such that $f(x)$ will never get closer to $c$ then distance $\epsilon$.

The wikipedia page on the ($\epsilon$, $\delta$)-definition is pretty good. Also, the picture from there may be helpful when you try to visualize $\delta$ and $\epsilon$ as distances.

delta, epsilon limit-definition visualization

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