How do you prove the statement lim as x approaches 2 for # (x^2 - 3x) = -2# using the epsilon and delta definition?

1 Answer
Oct 19, 2015

See the explanation, below.

Explanation:

Preliminary Analysis

We want to make #abs((x^2-3x)-(-2)) < epsilon#

By choosing #delta#, we control the maximum size of #abs(x-2)#

We now look at #abs((x^2-3x)-(-2))# keeping in mind what we control.

#abs((x^2-3x)-(-2)) = abs(x^2-3x+2)#

# = abs((x-1)(x-2))#

# = abs(x-1)abs(x-2)#

If we knew the size of #abs(x-1)#, we could choose #abs(x-2)# to make sure that the product is #< epsilon#

Let's start by making sure that #x# is a little close to #2#, say #abs(x-2) < 1# (#delta# will need to be at most #1#.)

If #abs(x-2) < 1#, then #-1 < x-2 < 1#, that is: #1 < x < 3#.

If follows that #0 < x-1 < 2#, so we have #abs(x-1) < 2#

Now if we ALSO make sure that #abs(x-2) < epsilon/2#,

then we will have #abs(x-1)abs(x-2) < 2abs(x-2) < 2 (epsilon/2) = epsilon#

Now we are ready to write the proof:

Proof

Given #epsilon > 0#, let #delta = min{1,epsilon/2}#.
(Note that #delta is positive.)

For every #x# that satisfies #0 < abs(x-2) < delta#
we have

(#abs(x-1) < 2# and also)

#abs((x^2-3x)-(-2)) = abs(x^2-3x+2)#

# = abs((x-1)(x-2))#

# = abs(x-1)abs(x-2)#

# < (2)(epsilon/2) = epsilon#

We have shown that for any #epsilon > 0# there is a #delta > 0# such that

if #0 < abs(x-2) < delta#, then #abs((x^2-3x)-(-2)) < epsilon#.

By the definition of limit, #lim_(xrarr2)(x^2-3x) = -2#.