The preliminary analysis is a bit long. If you just want to read the proof, scroll down.
Preliminary analysis
We want to show that #lim_(xrarr2)(x^2-7x+3) = -7#.
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^2-7x+3)))_(color(red)(f(x)) )-underbrace(color(blue)((-7)))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((2)))_color(green)(a))#
Look at the thing we want to make small. Rewrite this, looking for the thing we control.
#abs((x^2-7x+3)-(-7)) = abs(x^2-7x+10)#
# = abs((x-5)(x-2))#
# = abs(x-5)abs(x-2)#
And there's #abs(x-2)#, the thing we control
We can make #abs(x-5)abs(x-2)< epsilon# by making #abs(x-2) < (epsilon)/abs(x-5)#, 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 or equal to #1#, then
for every #x# with #abs(x-2) < delta#, we will have #abs(x-2) < 1#
which is true if and only if #-1 < x-2 < 1 #
which is true if and only if #1 < x < 3#
which, is ultimately equivalent to #-4 < x-5 < -2#.
Consequently: if #abs(x-2) < 1#, then #abs(x-5) < 4#
If we also make sure that #delta <= epsilon/4#, then we will have:
for all #x# with #abs(x-2) < delta# we have #abs((x-2)(x-5)) < delta * 4 <= epsilon/4 * 4 = epsilon#
So we will choose #delta = min{1, epsilon/4}#. (Any lesser #delta# would also work.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = min{1, epsilon/4}#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-2) < delta#, we have
#abs (x-5) < 4# and #abs(x-2) < epsilon/4#. So,
#abs((x^2-7x+3)-(-7)) = abs(x^2-7x+10)#
# = abs(x-5)abs(x-2)#
# < 4 * delta <= 4 * epsilon/4 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-2) < delta#, we have #abs((x^2-7x+3)-(-7)) < epsilon#
So, by the definition of limit, #lim_(xrarr2)(x^2-7x+3) = -7#.
