Using the Formal Definition

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Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

limx → √3 1/x² = 1/3

Accepted Answer

Welcome back, everyone. In this problem, we want to prove the limit statement that the limit of 4 divided by X2red as X approaches negative to equals 1 by choosing the correct value of delta in terms of epsilon that proves the given limit. A says Delta is the minimum value of 14th of Epsilon, B the minimum of 1.5 of Epsilon. C the minimum of 1 and Epsilon, and D, the minimum of 1 and 5 Epsilon. Now how are we going to prove the limit statement and choose the correct value of delta? What do we know about proving limit statements? Well, recall that based on the definition of the limit for every value of epsilon that's positive, OK. Then there will exist. A value of delta that's also positive such that such that if the difference between the the absolute value of the difference between X and the C that's the value that the limit approaches is greater than 0 but less than delta, then there exists. Or then we know that the absolute value of the function minus the limit will be less than epsilon. OK, so how can we apply that to our limit statement? Well, from our limit statement here we can tell that our function F of X is 4 divided by X2. C is equal to -2 since that's the value X is approaching and my limit equals 1. So if we apply the definition to a limit, OK, then that means it's going to say that for every value of epsilon that's greater than 0. There exists, OK. A value of delta that's greater than 0 such that OK, first. 0 is less than X minus -2 X minus C or the absolute value of X minus C. And if we simplify that, that's going to be the absolute value of X + 2, and that will be less than Delta, OK? So the absolute value of X + 2 is positive but less than Delta. Next, based on the definition, we also know that the absolute value of FF X that's 4 divided by X2. Minus 1, which is 1, is also going to be less than Epsilon. So no, for us to prove the given limit, we have to find delta in terms of Epsilon. In other words, we need to ask ourselves, how can we relate these two expressions or these two inequalities that can help us to solve or to figure out what value of delta we can use for this statement to be true. Well, let's start by looking at our absolute value of FF X minus L. Let's try to relate that to the absolute value of X + 2 somehow. No, if you take our absolute value here, OK. And if we were to combine it as one fraction, it's going to be the absolute value of 4 minus X2 divided by X2. Now if we use the difference of 2 squares here in our numerator, we could rewrite the absolute value of 4 minus X squared as 2 minus X multiplied by 2 + X or I'm going to write it as X + 2, OK? And all of that is being divided by X2. And no If we go ahead and simplify here, OK, then that means we could rewrite this, OK, if we think about it here. As the absolute value of 2 minus X multiplied by the absolute value of X + 2 divided by the absolute value of X2. Now, let's try to bone the absolute value of 2 minus X and X + 2, OK? Now, we know that since the limit is as X approaches -2, OK, then we can choose the value of delta that's less than or equal to 1, OK? So that When The absolute value of X + 2 is positive but less than Delta, then that means the absolute value of X + 2 is going to be less than 1. If this is true, this implies that X, OK, lies in the interval between parentheses -3, 1, closed parenthesis, OK? And in that case, that means, OK, that X + 2, or sorry, X minus 2 then is going to lie in the interval. Uh, between -5 and -3. If that's the case, Then that means the absolute value of X minus 2 is less than or equal to 5, OK. And the absolute value of X squared, OK. Which we know is just going to be X because X is going to give us a positive value whether X is positive or negative, so X2 is going to be greater than or equal to 1. Now if we combine these bones then. Then that means that the absolute value of 2 minus X multiplied by the absolute value of X plus 2. Divided by the absolute value of X2 is going to be less than or equal to 5 multiplied by the absolute value of X + 2. But remember this is the absolute value of 4 divided by X2 minus 1. So all of this is going to be less than epsilon. And thus if it is less than Epsilon to guarantee that 5 multiplied by the absolute value of X + 2 is less than Epsilon, then it is going to be sufficient. OK, it's gonna be sufficient. We can say it is sufficient to require that the absolute value of X + 2. Must be less than 50% of Epsilon, OK. No, based on that, what do, what does that tell us about our proof? Well, remember, we were trying to find delta in terms of Epsilon, and we know that the absolute value of X + 2 is less than Delta. So based on what we've done so far, we can tell then that in our proof we must satisfy the condition that absolute value of X + 2 is less than 1 to maintain our bones and that the absolute value of X + 2 is less than 1/5 of Epsilon, OK. Thus We can choose. A value of delta, that's going to be the minimum, OK. Of 1 and 1/5 of Epsilon. Thus, this is the value of delta that is going to make us be able to prove or limit statement and with this choice, then whenever the absolute value of X minus 2 is. Positive but less than delta, then we have the absolute value of 4 divided by X2 minus 1, less than or equal to 5 multiplied by the absolute value of X + 2, which is less than 5 multiplied by the multiplied by Epsilon divided by 5, which would be Epsilon. Thus, this is the value of delta we need. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

limx → √3 1/x² = 1/3

Key Concepts

Limit Definition

Continuous Functions

Rational Functions

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