Using the Formal Definition

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Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

lim x→1 1/x = 1

Accepted Answer

Welcome back, everyone. In this problem, we want to prove the limit statement that the limit of 1 divided by X as X approaches to equals 1/2 by choosing the correct value of delta in terms of epsilon that proves this limit. A says Delta is the minimum of 1 and 2 epsilon, B the minimum of 1.5 of epsilon. C that it is 2 epsilon, and D that it is Epsilon. Now how can we prove the limit statement and thus choose the correct value of delta? Well, recall that by the definition of the limit, it tells us that for every positive value of epsilon, OK. Then there exists a positive value of delta. There exists a delta that's greater than 0. Such that, OK. If 0 is less than the absolute value of X minus C, which is less than Delta, in other words, the difference between X and the value it approaching is less than delta but positive, then. The absolute value of our function F of X minus the limitl will be less than epsilon. Now if we look at our problem here, we can tell that C is equal to 2. OK. uh F of X is equal to 1 divided by X and L is equal to 1/2. So if we apply the limit definition to our problem, then we are saying, or we must show that for every epsilon that's greater than 0, OK, there exists. A value of delta greater than 0, such that. First, 0 is less than the absolute value of X minus 2, which is less than Delta. And second, that F of X minus L or the absolute value of FF X minus L, which is going to be 1 divided by X minus 12. Should be less than our value of Epsilon. So no, for us to choose a value of delta, let's see if we can try to express the absolute value of FFX minus L less than Epsilon in terms of the absolute value of X minus 2. We're trying to link both of them together. So how can we do that? Well, let's start with our expression here, OK? Now if we take the expression, the absolute value of 1 divided by x minus 1/2 and combine them into one fraction, it's going to be the absolute value of 2 minus x divided by 2 X. And no, we could rewrite that as the absolute value of x minus 2 because the absolute value of 2 minus X is the same as the absolute value of x minus 2. They both represent the difference. Divided by 2 multiplied by the absolute value of X, and this is what we know is going to be less than Epsilon. Now, in this problem, we want to control our denominator. OK? So let's start thinking about the bonds. Now, since the limit as X approaches, sorry, the limit is as X approaches 2, OK, then we want to choose a value of delta, OK. Choose a value of delta such that. The absolute value of X minus 2 is going to be less than 1, OK? If that's the case, Then that means XK lies in the interval. Let me write that here then. X will lie in the interval between 1 and 3. Which implies that the absolute value of X is going to be greater than or equal to 1. Since that's the case, OK, then that means the absolute value of 1 divided by X minus 1 divided by 2 is going to be less than or equal to the absolute value of X minus 2 divided by 2. Now what do we know here? Well, remember that this, let me move it down, uh, let me, let me carry it down to the next line. Remember that this expression is going to be less. OK, it's gonna be less than Epsilon. So to ensure that the absolute value of X minus 2 divided by 2 is less than Epsilon, then it is sufficient to require that the absolute value of X minus 2 be less than 2 Epsilon. Now if we think about it here, remember the absolute value of x minus 2 is less than delta. So since we also know that it's less than 1. And it's less than 2e, OK, to satisfy the condition, then in our proof we must satisfy that both the absolute value of X minus 2 is less than 1 to guarantee the bond and the absolute value of X and that it's less than 2 epsilon, which means that we can choose a value of delta. That's going to be the minimum, OK, of 1 and 2 epsilon. And know with just this choice, whenever the absolute value of X minus 2 is positive but less than Delta, then the absolute value of FF X minus L is going to be less than Epsilon. Thus, delta must be equal to the minimum of 1 and 2 epsilon. If we look back at our answer choices, that means A 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.

lim x→1 1/x = 1

Key Concepts

Limit Definition

Epsilon-Delta Proof

Continuity and Limits

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