Preliminary analysis
We want to show that #lim_(xrarr-1)((x+2)/(x-3)) = -1/4#.
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-3)))_(color(red)(f(x)) )-underbrace(color(blue)((-1/4)))_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)) = abs(x+1)#
Look at the thing we want to make small. Rewrite this, looking for the thing we control.
#abs((x+2)/(x-3) - (-1/4)) = abs((x+2)/(x-3)+1/4)#
# = abs(((4x+8)+(x-3))/(4(x-3)))= abs((5x+5)/(4(x-3))) = (abs5abs(x+1))/(abs(4)abs(x-3)) = (5abs(x+1))/(4abs(x-3))#
And there's #abs(x+1)#, the thing we control
We can make #(5abs(x+1))/(4abs(x-3)) < epsilon# by making #abs(x+1) < (4 epsilon)/(5abs(x+1))#, 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-(-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 #-2 < x < 0#
which, is ultimately equivalent to #-5 < x-3 < -3#.
Consequently: if #abs(x+1) < 1#, then #abs(x-3) < 5#
If we also make sure that #delta <= 4epsilon#, then we will have:
for all #x# with #abs(x+1) < delta# we have #abs((x+2)/(x-3)-(-1/4)) = (5abs(x+1))/(4abs(x-3)) < (5(4epsilon))/(4(5)) = epsilon#
So we will choose #delta = min{1, 4epsilon}#. (Any lesser #delta# would also work.)
Now we need to actually write up the proof:
Proof
Given #epsilon > 0#, choose #delta = min{1, 4epsilon}#. #" "# (note that #delta# is also positive).
Now for every #x# with #0 < abs(x-(-1)) < delta#, we have
#abs (x-3) < 5# and #abs(x+1) < 4epsilon#. So,
#abs((x+2)/(x-3) - (-1/4)) = abs((5x-5)/(x(x-3))) = (5abs(x+1))/(4abs(x-3)) <= (5abs(x+1))/(4(5)) < delta/4 <= (4epsilon)/4 = epsilon#
Therefore, with this choice of delta, whenever #0 < abs(x-(-1)) < delta#, we have #abs((x+2)/(x-3) - (-1/4)) < epsilon#
So, by the definition of limit, #lim_(xrarr-1)(x+2)/(x-3) = -1/4#.
