How do you prove the statement lim as x approaches -1.5 for # ((9-4x^2)/(3+2x))=6# using the epsilon and delta definition?

Question

Emilia Aguilar · Accepted Answer

You can't.

You can't prove it since it isn't 6. #lim_(x to -1.5) (9-4x)/(3+2)# does not exist. The limit from the left is #+oo# and the limit from the right is #-oo#.

Cameron Alexander · Answer

See the explanation.

Preliminary analysis We need to show that for every positive #epsilon#, there is a positive #delta# such that if #0 < abs(x-(-3/2)) < delta# we get #abs((9-4x^2)/(3+2x) - 6) < epsilon#. Note first that #abs(x-(-3/2)) = abs(x+3/2)# For every #x# other than #-3/2#, we have: #(9-4x^2)/(3+2x) = 3-2x#. So, for every #x# other than #-3/2#, we have: #abs((9-4x^2)/(3+2x) - 6)= abs((3-2x)-6)# # = abs(-2x-3) = abs((-4)(x+3/2)) = abs(-4)abs(x+3/2)#. # = 4abs(x+3/2)# We want this to be less than #epsilon# and we control, through #delta#, the size of #abs(x+3/2)# If we make #abs(x+3/2) < epsilon/4#, then we will have #4abs(x+3/2) < 4(epsilon/4) = epsilon# as desired. We are now prepared to write the evidence: Proof Given #epsilon > 0#, let #delta = epsilon/4#. Observe that this #delta# is also positive, as required. Now if #x# is chosen so that #0 < abs(x-(-3/2)) < delta#, the we have: #abs((9-4x^2)/(3+2x) - 6)= abs((3-2x)-6)# # = abs(-2x-3) = abs(-4)abs(x+3/2)#. # = 4abs(x+3/2) < 4delta = 4(epsilon/4) = epsilon# That is: if #0 < abs(x-(-3/2)) < delta#, then #abs((9-4x^2)/(3+2x) - 6) < epsilon#. So, by the definition of limit, #lim_(xrarr-1.5)((9-4x^2)/(3+2x)) = 6#

Axel Alvarez · Answer

undefined To prove the statement using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if the distance between x and -1.5 is less than delta, then the distance between the limit of the function and 6 is less than epsilon.

Let's start by finding the limit of the function as x approaches -1.5. We can simplify the function by factoring the numerator:

(9 - 4x^2) = (3 - 2x)(3 + 2x)

Now, we can rewrite the function as:

((3 - 2x)(3 + 2x))/(3 + 2x)

Canceling out the common factor of (3 + 2x), we get:

3 - 2x

To find the limit as x approaches -1.5, we substitute -1.5 into the simplified function:

3 - 2(-1.5) = 3 + 3 = 6

Now, let's proceed with the epsilon and delta proof. We want to show that for any epsilon greater than 0, there exists a delta greater than 0 such that if the distance between x and -1.5 is less than delta, then the distance between the limit of the function and 6 is less than epsilon.

Let epsilon be greater than 0. We need to find a delta such that if |x - (-1.5)| < delta, then |(3 - 2x) - 6| < epsilon.

Simplifying the inequality, we have:

|-2x + 3| < epsilon

Since we want to find a delta, we can assume that delta is less than 1 (to simplify calculations). Therefore, we can write:

|x + 1.5| < delta < 1

Now, we can manipulate the inequality to find a suitable delta:

|-2x + 3| < epsilon
|-2(x + 1.5)| < epsilon
2|x + 1.5| < epsilon
|x + 1.5| < epsilon/2

Since we assumed delta < 1, we can say that delta = min(1, epsilon/2).

Therefore, for any epsilon > 0, we can find a delta = min(1, epsilon/2) such that if |x - (-1.5)| < delta, then |(3 - 2x) - 6| < epsilon. This proves the statement using the epsilon and delta definition.