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(a) Estimate the value of $$ \lim_{x \to -\infty} \left( \sqrt{x^2 + x + 1} + x \right) $$ by graphing the function $ f(x) = \sqrt{x^2 + x + 1} + x $.

(b) Use a table of values of $ f(x) $ to guess the value of the limit.

(c) Prove that your guess is correct.

(a) From the graph of $f(x)=\sqrt{x^{2}+x+1}+x,$ we estimate the value of $\lim _{x \rightarrow-\infty} f(x)$ to be -0.5(b) From the table, we estimate the limit to be -0.5(c) $-\frac{1}{2}$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 6

Limits at Infinity: Horizontal Asymptotes

Limits

Derivatives

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This is problem number forty five of the Stuart Calculus Eighth addition Section two point six party estimate. The value of the limit is experts and I get of affinity of the square root of the quantity X squared plus x plus one plus x. My graph in the function F of X is equal to on the square root of the quantity exportable sex plus one complice X. So my graphing here we can see the grass of the function. Oh, but notable. Ah, point here towards very large negative number showing that the number is approximately negative zero point four nine three and the number It was slightly decreasing more and more as we went along. S so this is very close to about native one half. So the best estimate that we can get right now isn't this limit is approximately equal to negative one hour or is there a point five parties the table probably use of effects against the value of the limit. So if we were calling table values here and we see that we a cz, we increased the value the value of X to a more negative in negative, larger number at the value of the function itself perches Native Sierra Point fire. Until at this point, teams have equaled it. But of course, we're approaching the surprise. I haven't never reaching it, but we were able to confirm that are estimated from party is fairly accurate. And in part c, we're gonna prove that this estimate is absolutely correct. Analytically. So we're going to go ahead and workwith this limit here first going to multiply by the contra git to simplify this function a little further. So it's this quantity minus X divided by same quantity. Right More to plane by the contra. Get to the top and the bottom. This will give us it's limit. Tex Purchase negative infinity of Ah, this quantity X squared six. That's one my Mr X squared. That's what we get When we multiplied by the country and the denominator we saw the same term X squared plus x plus one Ah! And minus six. Okay. And the numerator We see that an X squared term can counsel and we're left with just experts. One. Want a little more supplication? Still, the limited sex purchase Negative Infinity, um generators X plus one in the denominator we're going to do is we're going to factor out and x squared from this radical. Which leaves us with one bus. I wanna rex this one over X squared. It's a little bigger. One plus one over X. This one Red X squared and then I'm in sex on the outside. Okay, Regarding this square root of X term since we're in the negative domain, we're going to take the following steps. The square root of X squared is equal to the upside Volume X And because excess listen zero we use a replacement. That devalue absolutely of X is equal to negative X again on ly in this region of our interest as experts Negative infinity. So we make that substitution Women his expressions negative infinity out of this quantity X plus one departed by negative X multiplied by this fraction or this radical or in plus one of Rex This one over x worried minus X. And our next step will be to divide by X so limited sex approaches. Thank you to infinity. Thanks a lot. About except one one of my exes. When Rex the radical that nominator doesn't change because this x is cancelled that just has a negative in the front. And then we have made a sextet, Rex, which is one and when exports is negative. Infinity dis fractions on Iraq's upper to Syria. Wanna wreck squared? Also purchase here and so we're left with is one place here in the numerator or one negative squirt of one minus one in the denominator which simplifies to make it a one or two or negative one, which confirms our estimates from parts and me.

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