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# Estimate the horizontal asymptote of the function$$f(x) = \frac{3x^3 + 500x^2}{x^3 + 500x^2 + 100x + 2000}$$by graphing $f$ for $-10 \le x \le 10$. Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy?

## so $y=3$ is a horizontal asymptote.The discrepancy can be explained by the choice of the viewing window. Try [-100,000,100,000] by [-1,4] to get a graph that lends credibility to our calculation that $y=3$ is a horizontal asymptote.

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this problem over fifty three of the Stuart Calculus eighth edition Section two point six estimate the horizontal, hasn't it? Of the function of alecks equals we execute those five hundred x word divided by the quantity X cube two plus five hundred x squared plus one hundred x plus two thousand by graphing F for X is between negative ten and ten. Then calculate the equation of the aspen took by evaluating the limit. How do you explain the discrepancy? So you take this function, we grabbed it first in this range from NATO ten to ten for axe. Here's an example of we're using a graphing and telling me to do this here in this window, we estimate the horizontal as, um took to be approximately ah y equals one. Seems as though the function approaches one as it gets closer and closer to ten and seems to level off at that point. So way, we say and re estimate that words don't ask until maybe why goals to one regarding the second part. We're going to calculate the limit now particular element as X approaches Infinity Ah, dysfunction three x cubed US five hundred x squared All right, Bye. Excuse me. So you want to x squared? Plus one thousand X? It was two thousand, and we're going to do as we take each term and divide by X cubed, which leaves us, which is three plus five hundred or X. Good bye. One plus five hundred or ex. It's one thousand over X squared class two thousand over X cubed and two properties of limits. We can take the limit of each term individually and his expert infinity. Each of these terms approaches hero, the ones that far over the form one over x Adar, where are is irrational value greater than zero and that leaves us with element equals two, three. And what that tells us is that our horizontal Lassiter is actually goes to three. So the discrepancy is there to drew planning. We estimated the horizontal hasn't turned out quite close to one where, as the actual Aspen to isa y equals three and the discrepancy has to do with there using a large enough graf or not understanding the scope of the graph anymore, because on their plant, the function could have increased way more, which it indeed does farther down to X axis in the negative and positive direction. And it indeed we'LL reach Ah! Oh, well, approach Quantico three whenever reach or echo three. Therefore, that is what explains the discrepancy.

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