Problem 35 Hard Difficulty

(a) For the limit $ \displaystyle \lim_{x \to 1}(x^3 + x + 1) = 3 $, use a graph to find a value of $ \delta $ that corresponds to $ \varepsilon = 0.4 $
(b) By using a computer algebra system to solve the cubic equation $ x^3 + x + 1 = 3 + \varepsilon $, find the largest possible value of $ \delta $ that works for any given $ \varepsilon > 0 $.
(c) Put $ \varepsilon = 0.4 $ in your answer to part (b) and compare with your answer to part (a).

Answer

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Video Transcript

This is problem number thirty five of this through a calculus eighth edition section two point four for the Limit Party for the limit limiters experts is one of the quantity. The function exit the third Power plus X plus one is equal to three. Use a graft to find a value of delta that corresponds to Absalon is equal to point four. Now, let's recall exactly the boy that Delta and Absalon play a role in our definition of a limit. We state that for we state that for and that's long greeted. Zero. There's Delta Creative Enzo such set the difference between X and A that absolutely of the difference is less than Delta. So if this is true, then they observe value. The function of minus the limit now is less than epsilon. So essentially the function the limit itself. So dysfunction in this limit value three, they differ only by value. Absalon. This corresponds to a difference between X and A in this case, sex in one. I know you don't don't. So we're gonna pull a graph of dysfunction. The red line represents the function which is execute X plus one. And here we've laid out and Absalon range of point four. So essentially we're approaching one X equals one that one. You know, the functions equals three and we and we also want to confirm that the limit is equals three. So as we saw here, the difference between the function and the limit I must be between arranging Peps lan. And since we're given point for, then we're gonna show at the point on the line out where the value is three plus epsilon So three plus point for that's three point four and also three minus phone for so the value of the function at two point six in this case three minus point four and the ex value's correspondent and are related to the delta The X values as a difference from the value that you're approaching this case it's one. Give us our delicate body. Therefore, you see that the corresponding X values are one point nine three for the upper limit one point nine three and then point eight nine one for the Lord. So we write party part A that we arrange. We range from pointing They won two one point nine three Rex and everyone represent this as ah X minus one. Subtract one from each term. Over here, we have point on nine. Three. Uh, and over here we have negative point one only on as we do in determining the Delta value that occupies the minimum of the absolute value Titties, Terminals. Because we want to make sure that we satisfy both Ah, both the upper limit of the absolute. So the absolute value, this is point one approximately point one after value. This is personal, and the minimum of those two is point for night. And that is what we choose our delta to be for party Ah, and part B by using a computer algebra system to solve the cubic equation executed plus X plus one is equal to three plus epsilon find the largest possible value delta that works for any given absolute greater than zero. And this is an exercise that you can use any algebra solver. You may have it handed you a graphing calculator. Or maybe line so use a graft from are an algebra solver and plug this in exactly on DH self For the value of X. In terms of Absalon, you should get a solution that similar to this, and at this time you may cause this explanation in order to chop down exactly this results. That's it. It's a pretty long, but it is exactly the result that is the solution to the cubic equation. And we see that it is a function of excellent. Uh, and it should work for any value of song that, just like them, with that being said, artsy asks us to input the body of absolute his point for and check our answer off this part using this equation from party and chicken with our answer for party. And if we take actually zero point for and plug it in here into each of the absolute terms that we see in this very long formula, uh, we actually will get a number of very consistent one party get that X is approximately moving one point o nine. That two seven two For Absalon, it's zero point four and this is very, very close to the value that we got from our graphene using autographing message in party. Therefore, we see that the delta value in this part that we get is this Xie minus one. So zero one zero nine two, seven two and we see that this agrees very well with the value we got in part eight and so this is just a more rigorous exact solution.