You need to prove that if 0 < |x-3| < epsilon/2 then |(2x-5) - 1| < epsilon.
(Precise Definition of Limit) For a function f defined in some open interval containing a (but not necessarily at a itself), we say
lim as x approaches a f(x)=L,
if given any number epsilion > ) there is another number delta > 0 , such that 0 < |x - a| < delta guarantees that
|f(x)-L|<epsilion
1. Symbolically find delta in terms of epsilion to prove the following limit is correct. (Use the above mentioned definition).
lim as x approches 3 (2x-5=1)