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

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Problem 18 Easy Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit and illustrate with a diagram like Figure 9.

$ \displaystyle \lim_{x \to -2}(3x + 5) = -1 $

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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

This is problem number eighteen of the Stuart Calculus. A fetish in Section two point four proved the statement using the absolute Delta definition of a limit and illustrate with the diagram like Figure nine the limit as X approaches negative, too. Of the function, three experts five is equal to negative one and their absolute out. The definition states that as long as we confined on absolute greater than zero for every Delta credit in zero system that exists under the conditions that if the difference between X and A absolutely use less than Delta, then the value of the difference between the function of its limits at that age it's less than that, son. Our function here again is three x plus five toe limit, given his negative one. The A given his thinking, too. So we're going to see what we can fearing about Ah relationship Pendleton Absalon. So we begin with the second inequality. Use our function. It's given three X plus five subject. Our known limit. What you think of one eso This minus one is plus one. It reduces to the absolute value the quantity three experts six is less than absolute. In fact, her out of three leaves us with absolute value of X plus two Western Absalon and then we divide by three. Now, at this point, we noticed that this would be the same or a dis pretty, very similar to this first inequality. Where is negative too. So this will become the absolute value of X Plus two. Listen, Delta. So we could make a statement and choose a Delta, um, and say that we can choose any Delta equal Teo Absalon over three and this would guarantee and prove our statement because we know of a delta. We know how to find a delta for every Absalon that allows for this limit to exist and equal need to run on our diagram to the right, we planted the function to expose five and in Vera in very similar fashion to fair one, we're going to show these Absalon Delta ranges for the Absalon corresponds to this y value. So it's going to be negative one minus sub saloon here in the lower and make it a one plus up slim. And then these correspondent too number two minus the altar on the left side. And they have two plus delta on the right side. And this diagram is consistent with their limit and the limit is proved

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