Using the Formal Definition

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Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

limx →4 (9 − x) = 5

Accepted Answer

Welcome back, everyone. In this problem, we want to prove the limit statement that the limit of 7 minus X as X approaches 2 equals 5. Which of the following choices for delta in terms of Epsilon correctly proves this limit? A says delta equals Epsilon, B a half of Epsilon, C 2 Epsilon, and the D Epsilon plus 1. Now, to prove this limit statement, we must show that for every positive value of Epsilon, there exists a positive value of delta, OK. So let's, let me write that down as we speak, OK. So we need to show. That for every positive value of Epsilon. OK. There exists a positive value of delta. Such that If X, the absolute value of X minus the value of limit is approaching, that is 2, is positive but less than Delta. Then The value of FFX 7 minus X, OK. The absolute, sorry, the absolute value of FFX 7 minus X minus or limit, which is 5, is going to be less than Epsilon. So how can we show this value of delta that exists and how can we figure out what that value of delta would be? Well, let's try to simplify our expression here for FFX minus L, the absolute value of FFX minus L. So if we take it here, so that's the absolute value of 7 minus X minus 5, then it's going to be equal to 7 minus 5, which is 2 minus X. That's the absolute value of 2 minus X, which we could rewrite as the absolute value of X minus 2. Now, why would we rewrite it as the absolute value of X minus 2? Well, if you think about it here, that's related to our definition for the absolute value of X minus 2 being less than delta but positive. OK. So since we know that F of X minus L is less than Epsilon, OK, and it eventually works out to the absolute value of X minus 2. And that means the absolute value of X minus 2, which is less than Epsilon. OK. Corresponds to where we say the absolute value of X minus 2 must be less than Delta. So since it's, it's, it's less than Epsilon and less than Delta, then it will hold the inequality will hold automatically if we set delta to Epsilon. OK? So if we said. Delta to Epsilon, OK. Then, If 0 is less than absolute value of X minus 2, which is less than Delta, then. 7 minus X minus 5, which equals the absolute value of x minus 2 will be less than Epsilon. Therefore, for this to hold true, then delta must be equal to Epsilon. 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.

limx →4 (9 − x) = 5

Key Concepts

Limit Definition

Epsilon-Delta Proof

Function Behavior Near a Point

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