# How do you prove that the limit of #((9-4x^2)/(3+2x))=6# as x approaches -1.5 using the epsilon delta proof?

Please see below. Warning: long answer due to explanatory analysis.

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

We have been asked to show that

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

Proving our L is correct -- Writing the proof

Proof:

Note

By signing up, you agree to our Terms of Service and Privacy Policy

To prove that the limit of ((9-4x^2)/(3+2x)) is 6 as x approaches -1.5 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 if 0 < |x - (-1.5)| < delta, then |((9-4x^2)/(3+2x)) - 6| < epsilon.

Let's proceed with the proof:

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

|((9-4x^2)/(3+2x)) - 6| = |((9-4x^2)/(3+2x)) - (6(3+2x)/(3+2x))| = |(9-4x^2 - 6(3+2x))/(3+2x)| = |(-12x^2 - 12x + 27)/(3+2x)|

Now, we can simplify the expression further:

|(-12x^2 - 12x + 27)/(3+2x)| = |(-3(4x^2 + 4x - 9))/(3+2x)| = |-3(2x-1)(2x+9)/(3+2x)|

Since we are interested in the limit as x approaches -1.5, we can substitute -1.5 into the expression:

|-3(2(-1.5)-1)(2(-1.5)+9)/(3+2(-1.5))| = |-3(-4)(6)/(3-3)| = |-72/0|

At this point, we can see that the denominator is zero, which means the limit does not exist. Therefore, we cannot prove that the limit of ((9-4x^2)/(3+2x)) is 6 as x approaches -1.5 using the epsilon-delta proof.

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- How do you use the Squeeze Theorem to find #lim xcos(1/x)# as x approaches zero?
- How do I use properties of limits to evaluate a limit?
- For what values of x, if any, does #f(x) = 1/((x-2)sin(pi+(8pi)/x) # have vertical asymptotes?
- If limit of #f(x)=3/2# and #g(x)=1/2# as #x->c#, what the limit of #f(x)/g(x)# as #x->c#?
- What is the limit of #( x^3 - 8 )/ (x-2)# as x approaches 2?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7