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Prove the statement using the $ \varepsilon $, $ …

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Problem 24 Easy Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to a} c = c $

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 4

The Precise Definition of a Limit

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Video Transcript

this problem Number twenty four of this tour Calculus eighth edition Section two point four Prove the statement using the absolute out the definition of a limit The limit is, experts say, of sea. Is he called to see you? Ah, the Epsilon Delta definition of limit is that for any absolute greater than zero there exists a delta greater than zero such that if the absolutely of the difference between exiting is less than a delta, then you have the value of the different between the function and l will be less than absolute. We'LL begin with the second limit and work from there are the second inequality. Second, inequality states that absolutely of the function In this case, the function is seen minus the limit. Tell him it is equal to seem This must be lesson Absalon, if we take C from CC represents a constant value to any content and it's the exact same constant zero and the absurd values here with zero and we get the statement that zeroes less than epsilon. Now what we should do first is we call Absalon is always greater than zero. That was one of our conditions and here and this. Working with the second inequality, we get a statement that is always true because of this case. Absolute skated in zero absolutely great isn't zero. Therefore, we can conclude any Delta choi. Any choice of Delta allows for our statement to be consistent, since regardless of any absolute that we choose, any Delta will be permissible because the statement will always be true and this limit always exists, regardless of choice of the telltale.

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