How do you use the epsilon delta definition to prove a limit exists?

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Answer 1 · 2015-08-16T09:38:20

It depends on the example, but basically you show

"for any $epsilon > 0 EE delta > 0 : ...$ "

in order to prove

" $AA epsilon > 0 EE delta > 0 : ...$ "

See an example in explanation.

Consider the function

$q(x) = { (1, "if x = 0"), (1/q, "if x = p/q for integers p, q in lowest terms"), (0, "if x is irrational") :}$

Then $q(x)$ is continuous at every irrational number and discontinuous at every rational number.

Let us show that $q(x)$ is continuous at any irrational number $alpha$ by showing that $lim_(x->alpha) q(x)$ exists and is zero.

Let $alpha$ be an irrational number and $epsilon > 0$

We need to show that:

$EE delta > 0 : AA x in (alpha-delta, alpha+delta), abs(q(x)-0) < epsilon$

Let $I = (alpha-1, alpha+1)$ so $alpha in I$

Let $N = ceil(1/epsilon) + 1$ , so $1/N < epsilon$ .

Let $S = { p/q : p, q in ZZ, 1 <= q <= N, "hcf"(p, q) = 1 } nn I$

Then $S$ is finite and for all $x in S$ we have $abs(x - alpha) > 0$ since $alpha$ is irrational.

Note that if $x in I and x !in S$ then $abs(q(x)) < 1/N$

Let $delta = min_(x in S) abs(x - alpha)$

Then since $S$ is finite, $delta > 0$ .

From our definition of $delta$ , if $x in (alpha-delta, alpha+delta)$ , then $x in I$ , $x !in S$ and $abs(q(x)) < 1/N$ .

So $AA x in (alpha-delta, alpha+delta), abs(q(x) - 0) < 1/N < epsilon$