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Problem 23 Easy Difficulty

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

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


Given $c>0,$ we need $\delta>0$ such that if $0<|x-a|<\delta$, then $|x-a|<c .$ So $\delta=\epsilon$ will work.


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Video Transcript

this problem. Number twenty three of the Stuart Calculus Ease Edition Section two point four. Prove this statement using the Epsilon Delta definition of a limit. The limit is expert is a of X is equal time. The AB signed out. The dimensional limit is summarized the right For every Absalon greater than zero there must be a delta greater than zero Such sort of the absolutely of the difference between X and aim Ms Liston daughter then the absolutely of the different between FX and own it must be lesson Absalon, if all the conditions are meant than the limit is a proven statement. So what we do is we're going to take Thesiger and Inequality and just use our function F which is in this case x minus l, which is a limit in this case. The limit is equal to a and said this less than Absalon. We also want to work with our first inequality. Experts say a being the value that this X is approaching. Should this problems, eh? His lesson, doctor, Comparing these two inequalities that we just determined we see that Delta equals Absalon is an appropriate choice. In order for these two to be consistent, right? If this first inequalities true, Delta's equal to Absalon. That makes this second equality, too. And we can choose any absolute now. And we have a Delta equal tips line. So there is a delta that agrees with any choice of song, and as such, all the conditions are meant and the statement and this limit is proof.