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) .
