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.
