Question

How do you use the epsilon delta definition of limit to prove that `lim_(x->5)(x-1)= 4` ?

Answer 1

Let us review the definition of a limit first.

Definition `lim_{x to a}f(x)=L` if
`forall epsilon > 0`, `exists delta > 0` s.t.
`0<|x-a|< delta Rightarrow |f(x)-L| < epsilon`

Let us now prove that `lim_{x to 5}(x-1)=4`.
(Note: `f(x)=x-1`, `a=5`, `L=4`)
Proof
`forall epsilon>0`, `exists delta=epsilon>0` s.t.
`0<|x-5|< delta Rightarrow |(x-1)-4|=|x-5|< delta= epsilon`.

Remark: The key is to find `delta` in terms of `epsilon`. In this particular proof above, we can set `delta=epsilon`.