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Problem 28 Easy Difficulty

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

$ \displaystyle \lim_{x \to -6^+} \sqrt[5]{6 + x} = 0 $


Choose $\delta=\varepsilon^{8}$ and then show that this implies that $|\sqrt[8]{6+x}-0|<\varepsilon$ for any choice of positive $\varepsilon$ and for any $x \in(-6,-6+\delta)$


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Video Transcript

This is problem number twenty eight of the Stuart Tactless eighth edition. Section two point four proved the statement using the Absalon Delta definition of a limit. Lim is X approaches negative six from the right of the function. The fifth root the quantities six plus X is equal to zero. Yep, sound out the definition of a limit states that for any absolute greater than zero, there must be a delta greater than zero such that if the absolute value of the difference between X and A is listen Delta. Then there's the value of the function. Minus the limit is less than absolute. To be in our problem, we're going to start with the second inequality and use our information here. Our fifth root of six plus six is our FX, and we're subtracting the algae limit, which is equal to zero and said to see Liston Absalon, we simplify this function. Man zero is just the function itself. And then here we raise both sides to the fifth power, such that the inside remains as a CZ. The inside of the fifth root, which is actually six and under, said we have e to the absolute to the fifth. Now, if we work with this other and equality, we have the absolutely X minus. Aye, aye. Being negative. Six lesson delta. And then we see that this is the same as the absolute. The quantity X plus six is. Listen, daughter, if we compared this inequality with the one we just found for Absalon, we see that a choice of Delta being equal to the to Absolut to the fifth power at this choice of Delta Ah is enough to prove our statement. We are We found a daughter for any Absalon. So as long as we have an absolute credit in zero, we can determine the daughter through this manner. And it confirms each of the inequalities that we needed to Ah, to satisfy the definition of the limit. And in such manner, we proved proven the statement that this limit is equal to zero.

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