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(a) Graph the function $$ f(x) = \frac{\sqrt{2x^2 + 1}}{3x - 5} $$How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits $$ \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} \hspace{5mm} \text{and} \hspace{5mm} \lim_{x \to -\infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} $$

(b) By calculating values of $ f(x) $, give numerical estimates of the limits in part (a).

(c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.]

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What this problem Number fifty four of this tour Calculus Safe Edition section two point six Party graft. The function F of X is equal to the square root of the quantity to X squared plus one. What about the quantity? Three X minus five. How many's horizontal and vertical asked meto see observed. And then we'll use the graft asked me to value these limits first, Let's take a look at the ground. This is thie graft function and we estimate there is one vertical as you do here approximately one and a half and we see there may be two separate horizontal aspirin, toots One maybe around one house positive one of and then another. Possibly it negative one half. And that's something we can I asked to meet here again The limits specifically, we're asking for those particular horizontal as toads. Here we have The limit is X approaches Infinity. That's the horizontal hasn't to here and sex purchase impunity. We see that dysfunction approaches approximately one half so well. All right, this here for now, approximately one half. And for this second limit as expert is negative Infinity. Ah, for the function we see that it approaches personal negative one half and again, we're just going to write thesis here to the side. OK, Herbie, calculating values of X give numerical estimates of the limits of party s. So what we do is we keep plugging in our general X values and for this function going towards infinity and larger and larger negative values towards the negative infinity. Here's a spreadsheet showing in the calculation of this function for a large, positive allies. And we see that function approaches value approximately equal to point four seven one four. Oh, fine. If we plug in Archer negative numbers towards negative infinity, we see that we approach the same limit. Just the negative. The opposite value. So four point four seven one four two four to some places seems like much better estimate. We're going to say that our estimates are now closer. Minus zero point four seven one four came Martine carefully. The exact values is limit in party, and they're asking, Did you get the same value or different values for these two limits? Okay, in order to calculate these limits exactly, we're going to use a bit of our, um Lim rules. Ah, one one one thing we could do. Take this function here, rewrite it as square root of X squared time to squared of two plus one over X squared. And this is just a factory and X squared numerator. Now it's important to realize that if X is positive, this value squared of X word eyes equal to X. If X is negative. Sequel to negative X That's what That again if X is positive, creative zero squared of X squared is positive. X. If X is less than zero, this term will be, well, two native X Um and this is important because we have to Domain's they were doing this limit on X course towards infinity is in the positive domain. So this functional be equal to X and for the other limit towards negative infinity dysfunction is equal to negative X on. That will change our answers slightly. Uh, for this positive limit, we'LL be able to reduce this to X. And if we divided by ex teach term we'LL hang out this limit two plus one or X squared, divided by three minus friend or ex And as its limit approaches infinity, the's terms go to zero. So our limit is actually equal to square root of two or three. And if we do it towards negative infinity, the second woman, remember, recall that this term becomes negative X and so are steps are approximately the same except with this negative in the front two plus one over X squared over three minus five over X. Each of these terms go to zero as experts Negative infinity on our final Ansari Cummings negative square, too over three. And if we were to calculate this exact value naked to score two to retreat, it is approximately equal to zero point for someone. For so in party with calculated the exact size of the limits Positive square in a two or three and they give us order to retreat. And those two are the horizontal incidents that we observed in our graph. Just it isn't very clear the way we crafted in this point. But if we expand closer to negative infinity and positive infinity, we should have confirmed that the horse on Jacinto are indeed are indeed at these eyes

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