How do you prove that the limit # (1-9x^2) / (1-3x) = 2# as x approaches 1/3 using the formal definition of a limit?

2 Answers
Nov 23, 2017

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Nov 25, 2017

Please see below.

Explanation:

Finding the values to use in 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)(1/3))color(red)((1-9x^2)/(1-3x) = color(blue)(2)#

So we want to make #abs(underbrace(color(red)((1-9x^2)/(1-3x) ))_(color(red)(f(x)) )-underbrace(color(blue)(2))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((1/3)))_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((1-9x^2)/(1-3x) - 2) = abs(((1-3x)(1+3x))/(1-3x) - 2)#

# = abs(((1+3x) - 2)#

#=abs(3x-1)#

Recall that we control the size of #abs(x-(1/3))# and if we factor a positive #3# out of the last expression we get

#=3abs(x-1/3)#

In order to make this less than #epsilon#, it suffices to make #abs(x-(1/3))# less than #epsi/3#

Writing the proof

Claim: #lim_(xrarr1/3)((1-9x^2)/(1-3x) ) = 2#

Proof:

Given #epsilon > 0#, choose #delta = epsilon/3#. (Note that #delta# is positive.)

Now if #0 < |x-(1/3)| < delta# then

#abs((1-9x^2)/(1-3x) -2) = abs(((1-3x)(1+3x))/(1-3x) - 2)#

# = abs(((1+3x) - 2)#

#=abs(3x-1)#

# = abs(3)abs(x-1/3)#

# = 3 abs(x-1/3)#

# < 3 delta#

# = 3 (epsi/3)#

# = epsilon#

We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(1/3)) < delta#, then #abs((1-9x^2)/(1-3x) - 2) < epsilon#.

So, by the definition of limit, we have #lim_(xrarr1/3)((1-9x^2)/(1-3x) ) = 2#.

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-1/3))#, then a choice of #x = 1/3# would result in an undefined expression. #abs((1-9x^2)/(1-3x) - 2) #.