How do you prove that the limit of #(x+2)/(x-3) = -1/4# as x approaches -1 using the epsilon delta proof?

Answer 1

See the section below.

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#.
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Answer 2

To prove that the limit of (x+2)/(x-3) is -1/4 as x approaches -1 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that whenever 0 < |x - (-1)| < delta, then |(x+2)/(x-3) - (-1/4)| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a suitable delta > 0.

First, let's simplify the expression |(x+2)/(x-3) - (-1/4)|:

|(x+2)/(x-3) + 1/4| = |(4(x+2) + (x-3))/(4(x-3))| = |(5x + 5)/(4x - 12)|

Now, we can proceed with the proof:

Choose delta = epsilon/20.

Assume that 0 < |x - (-1)| < delta.

Then, we have:

|x + 1| < delta

|x + 1| < epsilon/20

|x + 1| * |4x - 12| < (epsilon/20) * |4x - 12|

|(x + 1)(4x - 12)| < (epsilon/20) * |4x - 12|

|4x^2 - 8x - 12x + 24| < (epsilon/20) * |4x - 12|

|4x^2 - 20x + 24| < (epsilon/20) * |4x - 12|

4|x^2 - 5x + 6| < (epsilon/20) * |4x - 12|

4|x - 2||x - 3| < (epsilon/20) * |4x - 12|

Since we have assumed that 0 < |x - (-1)| < delta, we can also assume that |x - 2| < 3 and |x - 3| < 3.

Therefore, we have:

4|x - 2||x - 3| < 12|x - 2|

12|x - 2| < (epsilon/20) * |4x - 12|

12|x - 2| < (epsilon/20) * 4|x - 3|

3|x - 2| < (epsilon/5) * |x - 3|

Now, we can choose a suitable value for delta. Let's choose delta = min(1, epsilon/5).

If we assume that 0 < |x - (-1)| < delta, then we have:

|x - 2| < delta

|x - 2| < min(1, epsilon/5)

3|x - 2| < 3 * min(1, epsilon/5)

3|x - 2| < min(3, 3 * epsilon/5)

3|x - 2| < min(3, epsilon)

Therefore, we can conclude that for any given epsilon > 0, if we choose delta = min(1, epsilon/5), then whenever 0 < |x - (-1)| < delta, we have:

3|x - 2| < min(3, epsilon)

And since 3|x - 2| < min(3, epsilon), we can also conclude that:

|(x+2)/(x-3) - (-1/4)| < epsilon

Hence, the limit of (x+2)/(x-3) as x approaches -1 is -1/4, as proven using the epsilon-delta proof.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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