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# (a) How large do we have to take $x$ so that $1/\sqrt{x} < 0.0001$?(b) Take $r = \frac{1}{2}$ in Theorem 5, we have the statement $$\lim_{x \to \infty} \frac{1}{\sqrt{x}} = 0$$Prove this directly using Definition 7.

## $$\lim _{x \rightarrow \infty} \frac{1}{\sqrt{x}}=0$$

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this brown number seventy six of the Stuart Calculus eighth edition Section two point six Party How large do we have to take X so that one over the square root of X is less than point zero zero zero one? We'Ll solve this problem part of the problem to break clean Ah, one over discord of x less than ah one one one ten thousand. Our next step would be to take the reciprocal scored of X. We need to change the sign of the inequality greater than ten thousand and then Xmas speed greater than ah, we take the square We square both sides ten thousand squared or ten thousand squared is equal to one hundred million. So the answer to party X must be at least one hundred million in order for this function here to be less than one ten thousand party take our isn't one have Indira five, right? So this denominator here the extra the ex term has an exponent of one half, which is why it's screwed. We have the statement Limited's experts infinity of one over the square root of X equals zero proved this directly using definition seven. So definition seven states that if we restrict X to a certain value and, uh, then the difference between the function and its limit will be lesson certain error, certain accuracy, Absalon. And this needs to be true for any value of Epsilon chosen or any value. Then both of those need to correspond. And this is important because in order to confirm this limit, that is to be true for every absolute and such that the the difference between the function and limit right goes towards what is supposed to be, which is zero. So we can start with this second part the absolutely of the function one over the squared of X finest element zero Listen, value. Absalom, I swore was shipped ups on two positive values only for the moment, which means we can disregard this. Absolutely. Simon, listen, Absalon, and take this reciprocal of both sides. Sort of X is greater than one over Absalon and we can now square both sides. One squared over Absalon squared. Now we also noticed that this is related to our condition. Here, X is greater than the value. And so as long as we choose an end value equal to or greater than one over Absalon squared. Ah. Then we can confirm that this function will indeed be, um, the difference between the function. The limit will definitely be within its thiss accuracy of

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