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→5 1/x^2=1/25

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 1 divided by X squared as X approaches 7 is equal to 149th using the epsilon minus delta definition of the limit. For answer choices A says delta equals the minimum of 5 and 588 epsilon divided by 5. B the minimum of 1 and 5 epsilon divided by 588. The minimum of 1 and 588 divided by 5 epsilon and D the minimum of 1.588 epsilon divided by 5. Now, we need to select the correct relationship using the Epsilon minus delta definition of a limit. How can we do that? Well, recall this definition tells us that given the limit, OK. Of FFX As X approaches C, it is equal to L, OK. Then if epsilon is greater than 0, no matter how small, and this is just, uh, this just tells us how far the set of values are from F of X, then. There is a value of delta that's also greater than 0, OK. So let me write that properly here. There exists. A value of delta that's greater than 0, OK, such that, OK. 0 is going to be less than the absolute value of X minus C, which is less than Delta. And that would imply that there is an absolute value of the difference between F of the function and the limit. Sorry, that's going to be less than epsilon. So if we can prove this relationship, OK? And if we can show, assuming that this relationship is true, that Delta is related to Epsilon in some way or another, then we can find a relationship between Epsilon and Delta. So for our problem, let's let C be equal to 7, OK? And our limit L is 149th. Then 4, delta greater than 0. We know that we're going to have 0, less than X minus 7, less than Delta, OK. And for Epsilon, we're going to have the absolute value of FX, which is 1 divided by X minus our limit 1 divided by 49 to be less than Epsilon. Now, let's play our own with our value less than Epsilon for a bit to see if we can relate it to X minus 7 and eventually relate it to the absolute value of X minus 7 to get an idea of the relationship with Delta, OK. Now, we could rewrite 1 divided by X2 minus 1 divided by 49, OK? As 49 minus X2 divided by 49 X2. OK. Combine both of those fractions into one. Now we know that this is going to be equal to the absolute value of -149. OK, factoring out -149 of X2 minus 49 divided by X2, and we wrote it this way because by the difference of two squares we could rewrite X2 minus 49 as X minus 7 multiplied by X + 7. Likewise, we could write the absolute value of -149th, which is 149th to give us this expression as the same as the absolute value of 1 divided by X2 minus 149. Now, let's ask ourselves, how could we relate this to either X minus 7 or X + 7? Well, let's assume. That delta. is less than or equal to 1. OK. Notice that we want to control the distance of the absolute value of X minus 7. So let's focus on the X minus 7 term. Assuming that delta is less than or equal to 1, that means the absolute value of X minus 7 is going to be less than 1. And what that means then. Is that X minus 7 is going to be between -1 and 1. If that's the case, OK, then that means if we add 7 to both sides, then X is going to be between 6 and 8, OK. And We do this, we can assume this to keep X close to 7, and if we add 7 to both sides, that means 13 is going to be, sorry, X + 7 is going to be between 13 and 15, OK? Now here Given that 6. So, given that X is between 6 and 8, that means. X2 is going to be greater than 62, which is 36, OK. And We also know then that we can use this to further simplify our inequality because no, that means the absolute value of X minus 7 multiplied by X + 7 divided by 49 X2. OK. Is going to be less than the absolute value of our highest value for x + 715 multiplied by x minus 7 divided by 49 multiplied by x2 which we can we know that X2 is greater than 36. So no, with this idea to ensure, to ensure that the absolute value of 1 divided by X2 minus 1 divided by 49 is less than epsilon, OK, then we want, we want the absolute value of 15 multiplied by X minus 7 divided by 49 multiplied by 36 to be less than epsilon. OK? In this case, then we could rewrite this as the absolute value of X minus 7 as being less than 49 multiplied by 36 epsilon divided by 15, OK. And no if we factor out our common factor of 3 here, 336 goes 12, 3 into 15 goes 5 times. So that means the absolute value of X minus 7 is going to be less than 588 epsilon divided by 5. Now since we also said that Delta was less than or equal to 1, remember it told us that the absolute value of X minus 7 would be less than 1. Thus OK, we can set the set del sorry to be equal to the minimum value, OK, of either 1 or 588 epsilon divided by 5, whichever one is smaller, because remember, based on our definition of the limit. We know that the absolute value of x minus 7 would be less than delta. Thus, this would make this true. Now if we look back at our answer choices, then it tells us that D is the correct answer. Thanks 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→5 1/x^2=1/25

Key Concepts

Limit Definition

Epsilon-Delta Relationship

Continuous Functions

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