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

1 Answer
Wataru
Sep 7, 2014

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#.

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