How do you prove that the limit of #(4 + x - 3x^3) = 2 # as x approaches 1 using the epsilon delta proof?

1 Answer
Jan 16, 2018

See below.

Explanation:

Note that #(4+x-3x^3)-2 = -(3x^3-x-2) = -(x-1)(3x^2+3x+2)#

So #abs((4+x-3x^3)-2) = abs(x-1) * abs(3x^2+3x+2)#

Given #epsilon > 0#, let #delta = min{1, epsilon/20}# and note that #delta > 0#

If #0 < abs(x-1) < 1#
then #0 < x < 2# so that #2 < (3x^2+3x+2) < 20# and

#abs(3x^2+3x+2) < 20#

So, #abs((4+x-3x^3)-2) = abs(x-1) * abs(3x^2+3x+2)#

# < abs(x-1) * 20 < epsilon/20 * 20 = epsilon#

We have shown that for any positive #epsilon#, there is a positive #delta# such that

If #0 < abs(x-1) < 1# , then #abs((4+x-3x^3)-2) < epsilon#

So, by the definition of limit, #lim_(xrarr1)(4+x-3x^2) = 2#