Preliminary analysis
We want to show that #lim_(xrarr1)(x^2-x) = 0#.
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#.
So we want to make #abs(underbrace(color(red)((x^2-x)))_(color(red)(f(x)) )-underbrace(color(blue)(0))_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))_color(green)(a))#
Look at the thing we want to make small. Factor this, looking for the thing we control.
#abs((x^2-x)-0) = abs(x(x-1)) = absx abs(x-1)#
And there's #abs(x-1)#, the thing we control
We can make #absx abs(x-1) < epsilon# by making #abs(x-1) < epsilon/absx#, BUT we need a #delta# that is independent of #x#. Here's how we can work around that.
If we make sure that the #delta# we eventually choose is less than orequal to #1#, then
for every #x# with #abs(x-1) < delta#, we will have #abs(x-1) < 1#
which is true if and only if #-1 < x-1 < 1 #
which is true if and only if #0 < x < 2#.
Consequently: if #abs(x-1) < 1#, then #absx < 2#
If we also make sure that #delta <= epsilon/2#, then we will have:
for all #x# with #abs(x-1) < delta# we have #abs(x^2-x) = absx abs(x-1)<2(epsilon/2) = epsilon#
So we will choose #delta = min{1, epsilon/2}#. (Any lesser #delta# would also work.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = min{1, epsilon/2}#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-1) < delta#, we have
#absx < 2# and #abs(x-1) < epsilon/2#. So,
#abs((x^2-x) - 0) = absx abs(x-1)) <= 2abs(x-1) < 2delta <= 2 epsilon/2 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-1) < delta#, we have #abs((x^2-x) - 0) < epsilon#
So, by the definition of limit, #lim_(xrarr1)(x^2-x) = 0#.
