Question

How do you prove that the limit of `x-2 = 2` as x approaches 3 using the epsilon delta proof?

Answer 1

You can't prove it because it is false. `lim_(xrarr3)(x - 2) != 2`

Explanation:

You can prove thar `lim_(xrarr3)(x-2) = 1`

Proof

Let `epsilon > 0` be given. Choose `delta = epsilon`

Now for every `x` with `0 < abs(x-3) < delta`, we get

`abs(f(x)-L) = abs((x-2)-1) = abs(x-3) < delta = epsilon`.

We have shown that
for any `epsilon > 0`, there is a delta such that `0 < abs(x-2) < delta` implies `abs((x-2)-1) < epsilon`

Therefore, by the definition of limit, `lim_(xrarr3)(x-2) = 1`

Also, using a similar argument
we could prove that `lim_(xrarr3)(x-1) = 2`.

Sketch: again choose `delta = epsilon`

Then show that `0 < abs(x-3) < delta` implies `abs((x-1)-2) < epsilon`