How do you prove the statement lim as x approaches 1 for #(5x^2)=5# using the epsilon and delta definition?

1 Answer
Oct 25, 2015

See the explanation section below.

Explanation:

Preliminary Analysis

We need to make #abs(5x^2-5) < epsilon# by making #abs(x-1) < delta#. (We choose the #delta#.)

So we begin by examining #abs(5x^2-5)#

#abs(5x^2-5) = abs(5(x^2-1)) = abs5abs(x^2-1)#

# = 5 abs((x+1)(x-1)) = 5 abs(x+1) abs(x-1)#

We control, through our choice of #delta#, the maximum for #abs(x-1)#.

If we make sure that #delta# is less than a number we choose now, then we can also control the size of #abs(x+1)#.
(If we put a bound on the distance between #x# and #1#, the we also get a bound on the distance between #x# and #-1#)

Any number will do, but is might be easiest for purpose of illustration to pick a number we haven't use yet so we can keep track of it.

Let's make sure that #delta <= 2#.

If #abs(x-1) < 2#, then #-2 < x-1 < 2#.

So, adding 2 to each part, we get, #0 < x+1 < 4# and #abs(x+1) < 4#.

We want #5 abs(x+1) abs(x-1) < epsilon# and we plan to make sure that #abs(x+1) < 4#, so we now need

#5 abs(x+1) abs(x-1) #

#"which is" < 5(4)abs(x-1)#
# "which is"= 20 abs(x-1)#
#"we want this" < epsilon#

So let's make sure that, in addition to #delta <= 2#,

we also want #delta <= epsilon/20#

Now we are ready to write the proof.

Proof
Claim: #lim_(xrarr1)5x^2 = 5#

Given #epsilon > 0#, let #delta = min{2, epsilon/20}# (Note that #delta > 0# as required.)

If #x# is chosen so that #0 < abs(x-1) < delta# then

note first that #abs(x+1) < 4#
( #abs(x-1) < delta <= 2 rArr -2 < x-1 < 2 rArr 0 < x+1 < 4#).

Furthermore, for such #x#, we have

#abs(5x^2-5) = 5abs(x^2-1) = 5abs(x+1)abs(x-1)#

# < 5(4) abs(x-1) = 20abs(x-1)#

# < 20(epsilon/20) = epsilon#.

That is: for #0 < abs(x-1< delta#, we have #abs(5x^2-5) < epsilon.

Therefore, by the definition on limit,

# lim_(xrarr1)5x^2 = 5#