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(a) How large do we have to take $ x $ so that $ 1/x^2 < 0.0001 $?

(b) Take $ r = 2 $ in Theorem 5, we have the statement $$ \lim_{x \to \infty} \frac{1}{x^2} = 0 $$

Prove this directly using Definition 7.

(a) $1 / x^{2}<0.0001 \Leftrightarrow x^{2}>1 / 0.0001=10000 \Leftrightarrow x>100 \quad(x>0)$

(b) If $\varepsilon>0$ is given, then $1 / x^{2}<\varepsilon \Leftrightarrow x^{2}>1 / \varepsilon \Leftrightarrow x>1 / \sqrt{\varepsilon}$. Let $N=1 / \sqrt{\varepsilon}$

Then $x>N \Rightarrow x>\frac{1}{\sqrt{\varepsilon}} \Rightarrow\left|\frac{1}{x^{2}}-0\right|=\frac{1}{x^{2}}<\varepsilon,$ so $\lim _{x \rightarrow \infty} \frac{1}{x^{2}}=0$

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this problem Number seventy five of this tour. Calculus. Eighth edition, Section two point six, part eight. How large do we have to take X so that one over X squared is less than zero point zero zero zero one for this part? We're going ahead and going to solve this outbreak. Lee s. So when is this true? One over X squared. Less than point zero zero there. One that's the same Assane one over X squared. Ah, less than ten thousand or one over ten thousand. Now our next step will be to take the reciprocal of Kotite's changing this the direction of the inequality. Ah, and finally we take a square. So in this case, we need to take X as large as one hundred. In order for us to confirm that one over X squared is less than a one one one ten thousand X must be larger than one hundred. Prepare B. Yeah, according to deer and five we take, our is equal to two. So that means the exponent on the X and the denominator is too. We have this statement the limit as ex purchase infinity of the function one over X squared is equal to zero. Prove this circle using definition seven definition seven says if X is greater than the value and then the distance or the difference between the function on the limit will be less than a certain no margin. A certain error Absalon came. So we need to be able to confirm this is true for any choice of Absalon that we can find a certain end value s so we can proceed and use what function we have here. One over x squared, the limit l which is zero in this case less than Absalon. We're trying to figure out maybe what absolute should be. We're going to restrict Absalon to only positive values, which means that can disregard the absolutely sign. Ah, we can take the reciprocal X squared greater than Absalon and ah, or ex ex Greg greater than one over. Absolutely. And then finally the square root ex greater than one over the square root of Absalom. At this point, this is what satisfies this second part. The absolute value, the quantity effects, F X minus l being less in a quantity. Absalon. We also see that this has to be true for X is greater than a certain number. And so here we conclude that if we choose and to be at least one over the square root of Absalon Ah, then that confirms the statement. So for any Absalon, we can choose ups on to be one and choose Epsilon to be point one point zero zero one every single time. Well, we will be able to choose an appropriate and value such that this statement is true and according to definition seven, because that is true. This limit is in fact, true that the limited his expressions affinity of one over X squared is equal to zero.