Question

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

Answer 1

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.

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`