How do you prove that the limit of #((9-4x^2)/(3+2x))=6# as x approaches -1.5 using the epsilon delta proof?

Question

Emilia Aguilar · Accepted Answer

Please see below. Warning: long answer due to explanatory analysis.

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself. Finding the proof By definition, #lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)# if and only if for every #epsilon > 0#, there is a #delta > 0# such that: for all #x#, #" "# if #0 < abs(x-color(green)(a)) < delta#, then #abs(color(red)(f(x))-color(blue)(L)) < epsilon#. We have been asked to show that #lim_(xrarrcolor(green)(-1.5))color(red)((9-4x^2)/(3+2x) = color(blue)(6)# So we want to make #abs(underbrace(color(red)((9-4x^2)/(3+2x) ))_(color(red)(f(x)) )-underbrace(color(blue)(6))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((-1.5)))_color(green)(a))# We want: #abs((9-4x^2)/(3+2x) - 6) < epsilon# Look at the thing we want to make small. Rewrite this, looking for the thing we control. #abs((9-4x^2)/(3+2x) - 6) = abs(((3-2x)(3+2x))/(3+2x) - 6)# # = abs(((3-2x) - 6)# #=abs(-2x-3)# Recall that we control the size of #abs(x-(-1.5)) = abs(x-(-3/2))# I see #abs(x+3# which is the same as #abs(x-(-3))# so let's 'factor out #-2#. # = abs(-2)abs(x+3/2)# # = 2 abs(x-(-3/2))# In order to make this less than #epsilon#, it suffices to make #abs(x-(-1.5))# less than #epsi/2# Proving our L is correct -- Writing the proof Claim: #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6# Proof: Given #epsilon > 0#, choose #delta = epsilon/2#. (Note that #delta# is positive.) Now if #0 < |x-(-1.5)|=|x-(-3/2)|< delta# then #abs((9-4x^2)/(3+2x) -6) = abs(((3-2x)(3+2x))/(3+2x) - 6)# # = abs(((3-2x) - 6)# #=abs(-2x-3)# # = abs(-2)abs(x+3/2)# # = 2 abs(x-(-3/2))# # < 2 delta# # = 2 (epsi/2)# # = epsilon# We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(-1.5)) < delta#, then #abs((9-4x^2)/(3+2x) -6) < epsilon#. So, by the definition of limit, we have #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#. Note This is an example of a limit in which the strict inequality #0 < abs(x-delta)# is very important. If we allowed #0 = abs(x-(-1.5))#, then a choice of #x = -1.5# would result in an undefined expression. #abs((9-4x^2)/(3+2x) -6) #.

Emma Aldridge · Answer

undefined To prove that the limit of ((9-4x^2)/(3+2x)) is 6 as x approaches -1.5 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - (-1.5)| < delta, then |((9-4x^2)/(3+2x)) - 6| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a suitable delta > 0. |((9-4x^2)/(3+2x)) - 6| = |((9-4x^2)/(3+2x)) - (6(3+2x)/(3+2x))| = |(9-4x^2 - 6(3+2x))/(3+2x)| = |(-12x^2 - 12x + 27)/(3+2x)| Now, we can simplify the expression further: |(-12x^2 - 12x + 27)/(3+2x)| = |(-3(4x^2 + 4x - 9))/(3+2x)| = |-3(2x-1)(2x+9)/(3+2x)| Since we are interested in the limit as x approaches -1.5, we can substitute -1.5 into the expression: |-3(2(-1.5)-1)(2(-1.5)+9)/(3+2(-1.5))| = |-3(-4)(6)/(3-3)| = |-72/0| At this point, we can see that the denominator is zero, which means the limit does not exist. Therefore, we cannot prove that the limit of ((9-4x^2)/(3+2x)) is 6 as x approaches -1.5 using the epsilon-delta proof.