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Problem 73 Hard Difficulty

For the limit $$ \lim_{x \to -\infty} \frac{1 - 3x}{\sqrt{x^2 + 1}} = 3 $$ illustrate Definition 8 by finding values of $ N $ that correspond to $ \varepsilon = 0.1 $ and $ \varepsilon = 0.05 $.

Answer

$N=-19$

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Video Transcript

this problem. Number seventy three of the Stuart Calculus. A fetish in section two point six for the limit. Lim is X approaches Negative infinity one minute, three x divided by the quantity X squared plus one equal to three illustrate definition. Eight. Okay, Finding Malley's end the correspondent to absolutely eyes equal to zero point one an absence he called. Is there a point zero five? And his problem has to do with guaranteeing that the function and its limit the distance between the two remains less than a certain quantity? Absolutely. And looking for and X value. Ah guarantees that this will be true. Our problem our function is this one minus three x over this crude of the quantity x four plus one on the limit is three. So Ah, oui is a graphing tool to compare these two and make sure that the distance between them is less than absolute. So if we take a look at the function plotted and red here, we see that the function going towards negative affinity increases and Jim's past the value of three and then as it a purchase Negative infinity. It approaches three as X goes towards negative infinity, but it never reaches three. So in this case it is approaching three that will never, never go less than three. So if we are interested in the distance between the function room and the limit being less than point one, then we're interested in where the function suddenly becomes less than point one within three. So a distance of point one above three, which is three point one. So at this point, we're decreasing X. We're going more negative in the extraction. Once we had to find you of eight point one, we're now within point one of element point one of three. And for this reason, negative eight point one is the value. Then anyway, only listen. That guarantees that the function is with zero point one. I'm the limit. Using a much more strict rally, your absalon point zero five. We see that we're going to need a much larger negative value. And as we go along dysfunction, we can trace the function too. Value of approximately negative eighteen point four, at which point we know that it is within zero point zero five are the limit, which is three, and anything less than that. We'LL definitely be within their point zero five. So the only event for the second Absalom is negative. Hey, teen point four and those are answers for this problem.