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Prove the statement using the $ \varepsilon $, $ …

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Problem 22 Medium Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to -1.5}\frac{9 - 4x^2}{3 + 2x} = 6 $

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

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Video Transcript

prove this statement. This is problem number twenty two of these to a cactus. In addition, Section two point four through the statement using the Absalon Delta definition of the limit, the limit is experts is negative one point five of the quantity and I'm minus four x squared, divided by the quantity three plus two X is equal to six. And here to the right, we have a summary of our absolute out. The definition of a limit is limited. It perches as a CZ expert initiative. One point five of this function is equal to six as long as there is an absolute great in zero for every Absalon great. And so that there is adulterated Terrel such that if the absolutely the difference between X and A in this case make it a one point five is lesson doctor, then they absolutely the function and the limit. That difference of the absolutely of that difference is less than Absalom. So we begin with our second equality and right the function nine minus for X squared. That quantity, divided by the quantity three plus two eggs minus the limit the limit is equal to six is less than absolute Absalon. We want to simplify distraction. Possibly if that helps simplify out of this further. We noticed that this is a difference of squares. This is three squared. This is to X squared. So we split it up as three plus two x and three minus two X denominator remains three plus two x six. Here we noticed that in the denominator, ACS is not allowed earthy value of negative one point five is not a little mean. And we you choose to assume that X only a purchasing at one point five and does not equal one point five. And since this is true, then these two Kim council thes two terms leaving us with three minutes to X minus six. If we're right that combining three and then ninety six, we get negative two X minus three Less than absolute Here we confected out. See, we can factor out or divided, divided by negative too, Charlie. And pull that out. So native to give us this X plus three or two listing Absalon notice that this is appropriate because they have to multiply by X Gives this thing up to x ray. Want to play by three over two times this number three This absolutely of the negative, too, is the same man's the absolutely positive, too. And that, too, can be divided on both sides and our final statement as X plus three. House is less than absolute over too. And we make the observation that air first and equality is X minus a, which is Ah, negative one point five. So, plus one point five must be lesson. Delta one point five is another form of the fraction three halves. And we make observation that this inequality and the one we found her previously are almost equivalent on their equivalent for Delta. He cools to Absalon over too. And here is our relationship between Delta Epsilon. Essentially, we have proven that there exists a delta greater than zero for every apps on agree with Theo. And that was thie proof for this limit. Through this absolute off the definition of limited that as long as Delta is equal to Absalon or two a cz an example, then this statement is proved

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