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Problem 26 Easy Difficulty

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

$ \displaystyle \lim_{x \to 0} x^3 = 0 $

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Video Transcript

this problem. Number twenty six of this tour. Calculus s edition, Section two point four. Prove the statement using the Absalon Delta definition of eleven. Tell him it is experts. Zero x cubed. Is he called a zero? The Epsilon Delta definition of element states that this limit is true and is equal to zero as long as there's an absalon to greater than zero that for every Absalon greater than zero, you find a delta greater than zero such that if the value of the difference between X and a his lesson delta, then the asset value of the difference being the function and the limit l is less than Absalon. We'LL begin with the second inequality and specifically playing our function of interest are function here is X cubed minus still amid which is equal to zero must be. Listen, Absalon, if we simplify this just a bit, get the cube root. And if we take the cube root on both science, we should get this statement here. Be absolutely vexes. Listen, the cube root of Absalon working with this first inequality the absolutely of X minus A. In our case, a is zero. That's what X is approaching is less than Delta should be. Zero. And we if we simplify experience zero, it just gives this up. So I vexes. Listen, Delta. And then at this point, we can compare these two inequalities and see that an appropriate choice of Delta would be dealt is equal to the cube root of Absalon. And this relationship itself proves that there is a delta created zero for any absolute zero as long as it's a plan is greater than zero that satisfies all these inequalities and statements as satisfies the definition of a limit. The excellent if not the definition of a limit which proves our statement proves the limit exists and is he cool to zero?