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Results for this submissionEnteredAnswer Previewsin(4)4*sin(X/4)x(it IodThe answer above is NOT correct:point) Evaluate the limit using Theorem 2 as necessary: lim ...

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Results for this submissionEnteredAnswer Previewsin(4)4*sin(X/4)x(it IodThe answer above is NOT correct:point) Evaluate the limit using Theorem 2 as necessary: lim st (4sin(x/4))xPreview My AnswersSubmit AnswersYour score was recorded. You have attempted this problem 3 times You received & score Of 0% for this attempt Your overall recorded score is 0%_ You have unlimited attempts remaining-Email instructor

Results for this submission Entered Answer Preview sin(4) 4*sin(X/4)x (it Iod The answer above is NOT correct: point) Evaluate the limit using Theorem 2 as necessary: lim st (4sin(x/4))x Preview My Answers Submit Answers Your score was recorded. You have attempted this problem 3 times You received & score Of 0% for this attempt Your overall recorded score is 0%_ You have unlimited attempts remaining- Email instructor



Answers

Find the limit and confirm your answer using the Squeeze Theorem. $$ \lim _{x \rightarrow \infty} \frac{\sin \left(x^{2}\right)}{x} $$

Yeah, for this problem we are asked to find the limit as X approaches negative infinity of sine of X divided by X. Using the squeeze zero. So the first thing that we can do is we know that sine of X is going to be between or less than or equal to negative one. Or rather it's going to be greater than or equal to one and it's going to be less than or equal to positive all. So that means then that sign effects over X is going to be between negative one over X and positive one over X. Which then means that if we take the limit as X approaches negative infinity that this inequality is going to be maintained. So actually I'll be careful here, we are still taking a limit. No, X approaches negative infinity of positive one over X. So when we take the limit as X approaches negative infinity of negative one over X we get zero that's less than or equal to the limit. As X approaches negative affinity of sine of X over X which in turn is going to be less than or equal to the limit as X approaches negative infinity of one over X which is again zero, which tells us that the lesson and greater than signs are not useful because we have this chain which suggests that the only possible way of making that inequality true is if the limit as X approaches negative infinity of sine X over X is equal to zero

For this problem, we are asked to find the limit and confirmed the answer using the squeeze theorem for the function one minus coast of X over X squared where we want to take the limit as X approaches infinity. Now, what we know about the function that's happening here is that with one minus coast of X? Well, there's not one single value that that will be approaching, but if we consider the absolute value of one minus cosine of X, well, we can consider coast of X that is going to be between negative one and positive one. So if we consider doing one minus coast of X, then we'll have actually I'll just write it out this way here For between 1- Coast of X. That will be then between Uh positive to when kovacs equals negative one And zero when Kovacs one. So that would tell us then that we should expect that as X grows very large, we'll have I'll write down sort of informally the limit actual. Just write down as X approaches infinity. X squared we expect will be much much much greater Then 1- Kovacs. So that would suggest to us then that our limit as X approaches infinity of one minus coast of X over X squared Is going to equal zero. Now we can check our answer because we know then that we are going to be greater than just zero because the minimum value of one minus coast back zero zero is going to be less than the limit as or less than or equal. I should say that should have been less than or equal to up there. Less than or equal to the limit as X approaches infinity of one minus cosine of X over X squared, which in turn is going to be less than or equal to the limit as X approaches infinity of two over X. That also should have been an X squared, their two over X squared. So when we take the limit as X approaches infinity of two over X squared, we can very clearly see that that will be equal to zero. So we get zero is less than or equal to our original limit, which must be then less than or equal to zero. Which means that the only possible answer is that the limit as X approaches infinity of one minus coast of X over X squared Is going to equal zero.

For this problem, we are asked to find the limit and confirmed the answer using the squeeze theorem for the function one minus coast of X over X squared where we want to take the limit as X approaches infinity. Now, what we know about the function that's happening here is that with one minus coast of X? Well, there's not one single value that that will be approaching, but if we consider the absolute value of one minus cosine of X, well, we can consider coast of X that is going to be between negative one and positive one. So if we consider doing one minus coast of X, then we'll have actually I'll just write it out this way here For between 1- Coast of X. That will be then between Uh positive to when kovacs equals negative one And zero when Kovacs one. So that would tell us then that we should expect that as X grows very large, we'll have I'll write down sort of informally the limit actual. Just write down as X approaches infinity. X squared we expect will be much much much greater Then 1- Kovacs. So that would suggest to us then that our limit as X approaches infinity of one minus coast of X over X squared Is going to equal zero. Now we can check our answer because we know then that we are going to be greater than just zero because the minimum value of one minus coast back zero zero is going to be less than the limit as or less than or equal. I should say that should have been less than or equal to up there. Less than or equal to the limit as X approaches infinity of one minus cosine of X over X squared, which in turn is going to be less than or equal to the limit as X approaches infinity of two over X. That also should have been an X squared, their two over X squared. So when we take the limit as X approaches infinity of two over X squared, we can very clearly see that that will be equal to zero. So we get zero is less than or equal to our original limit, which must be then less than or equal to zero. Which means that the only possible answer is that the limit as X approaches infinity of one minus coast of X over X squared Is going to equal zero.

So in this problem, we're going Teoh evaluate limits without using low petals rule. So we have signed over tan and let's go ahead and simple and rewrite this using the definition of tan that sign over co sign. Now notice that the science cancel So we're just left with co sign co sign at zero is one. That's our answer now for Part two. The limit goes to one and looks like yet, or we were going to wants a factor, so we have X minus one X plus one on the bottom X minus one X squared plus X plus one and looks like the X minus ones canceled. Explosive one X squared plus X plus one on the bottom. That's two thirds that's it.


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