Use the precise definition of a limit to prove the following l...

Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→3 x^3=27

Accepted Answer

Welcome back, everyone. In this problem, we want to select the correct relationship between Epsilon and Delta to prove that the limit of X as X approaches 2 equals 8 using the Epsilon minus delta definition of a limit. A says Delta is the minimum of 1 and epsilon divided by 19. B says it's the minimum of 1 and epsilon divided by 17. The minimum of 5 and Epsilon divided by 19. And the maximum of 1 and epsilon divided by 19. Now, how can we select the right relationship to prove this limit using the Epsilon minus delta definition of a limit? Well, first, let's think about that definition. Recall what it basically means is that if we are given the limit of F of X, OK, as X approaches C to be equal to L, OK. Then If Epsilon is greater than 0, that means a very small value that's close by to F of X, then there exists, OK, a value of delta that's greater than 0. That is a value that is close to or or value C, OK. Such that Such that If 0, OK, is less than or equal to the absolute value of X minus C, which is less than delta, it implies that the absolute value of the difference between the function and the limit is going to be less than epsilon. OK? So, essentially, if we are going to prove our limit and find a relationship that can prove the limit, we'll have to try to express or try to write our terms here. OK, I try to write our values here for Delta and Epsilon in the same way, and that will help us to eventually figure out what the relationship will be, OK? So for our problem. We can let C be equal to 2, OK. And L based on our problem statement to be equal to 8, OK. And in that case, or a value of delta. It's going to be that 0 is less than absolute value of x minus 2, which is less than delta, and for epsilon greater than 0, it's going to be that the absolute value of the function X cubed minus the limit 8 is going to be less than epsilon. Now is there some way that we can get the absolute value of X minus 8 to look like the absolute value of X minus 2? Well, we know that we could rewrite Xcubed minus 8. As Minus 2, OK, multiplied by X2 + 2 X + 4, OK? And now in this case, Then we could rewrite this as the product of the absolute value of X minus 2 and the absolute value of X2 + 2 X + 4. Now, let's assume, OK. Let's assume that. Delta is less than or equal to one as a very small value, OK. Then in that case that would mean the absolute value of x minus 2 would be less than 1. And if the absolute value of X minus 2 is less than 1, that implies that x minus 2 is between -1 and 1, which then tells us that X is going to be between -1 + 2 or 1 and 1 + 2 or 3, OK. So in other words, if Delta is less than or equal to 1, then X is going to be between 1 and 3. Now, under this assumption, we can estimate, OK, we can estimate X2, OK, or the absolute value of X2 + 2 X + 4. By finding its maximum, OK, in the interval. Because if we find its maximum in its interval, we know that that's the highest possible value that it could be, OK. And if we find the highest possible value it could be, then we, we can tell the values that it's going to be less than. OK. So in other words, by estimating that we're going to get X2 + 2 X + 4. To be less than 3 squared not replacing X with 3. Plus 2 multiplied by 3 + +43 squared is +92 multiplied by 3 is 6, and 9 + 6 + 4 equals 19, OK? So in that case, This means that. This value That we or this way that we've been able to rewrite X cubed minus 8, and there was the absolute value of X minus 2 multiplied by the absolute value of X2 + 2 X + 4, that's less than Epsilon can also be rewritten as 19 multiplied by the absolute value of X minus 2 less than Epsilon. In other words, we know that this expression must be less than Epsilon, OK? And in that case that tells us then that the absolute value of X minus 2 must be less than Epsilon divided by 190, eh. What does this mean? Well, no, OK, that means the absolute value of X minus 2 must be less than 1, or the absolute value of X minus 2 must be equal to epsilon divided by 19. And remember the absolute value of X minus 2 is less than delta, OK? So what this means then. Is that delta is going to be equal to the minimum value of either one. OK. R epsilon divided by 19. If we look back at our answer choices, that tells us then that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.
lim x→3 x^3=27

Key Concepts

Limit Definition

Epsilon-Delta Relationship

Polynomial Functions

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