We are to prove that no matter how small a tolerance e > 0 we desire for
(5-2x) to be within -1, we will always be able to find at least one value
for d > 0 such that whenever x is within d of 3, that (5-2x) will always be
within e of -1.
Suppose we are given e > 0 and desire to prove that there exists a value of
d > 0 such that |f(x)-(-1)| < e. We look at the absolute value of the
difference between the (5-2x) and -1
|(5-2x)-(-1)| = |5-2x+1| = |6-2x| = 2|3-x| = 2|x-3| we want to be < e
So the (5-2x) will be within e of -3 if 2|x-3| < e, which is to say
when |x-3| < e/2, so if we choose any d < e/2, then when |x-3| < d, that is,
when x is within d of 3, (5-2x) is within e of 3.
So given e > 0 |(5-2x)-(-1)| < e whenever d < e/2.
That proves that
Edwin
