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Question 1 (5 Points) State the Fundamental Theorem of Calculus (both parts)_Question 2.Points) Using 'Hopital's Rule student writeslitu T+ |liuyDo you ag...

Question

Question 1 (5 Points) State the Fundamental Theorem of Calculus (both parts)_Question 2.Points) Using 'Hopital's Rule student writeslitu T+ |liuyDo you agree with this student? Explain why or why not. If you do not Jgree, find the correct limit. Justify completely:

Question 1 (5 Points) State the Fundamental Theorem of Calculus (both parts)_ Question 2. Points) Using 'Hopital's Rule student writes litu T+ | liuy Do you agree with this student? Explain why or why not. If you do not Jgree, find the correct limit. Justify completely:



Answers

$1-38=$ Find the limit. Use l'Hospital's Rule where appropriate.
If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.
$$\lim _{x \rightarrow 0} \cot 2 x \sin 6 x$$

So part A. We have the limit as X approaches infinity of f of X plus x signed of X, divided by X squared minus four. So we divide the numerator and denominator expression by X square. We have thistle is equal to one over ax plus the sign of X divided by X all divided by one minus four over X squared. So this is equal to zero plus zero divided by one month Syria, which is equal to zero. Next for part B, we have the limit as X approaches infinity limit as X approaches infinity of f of X after lex, which is, um X plus X sign of X. And this is equal to infinity, plus infinity. So that's just infinity. And if we have the limit as X approaches infinity of G of axe, which is equal to X squared minus four, which is equal to infinity minus four, which is just equal to infinity. Finally, we have part C, which is if we take the limit as X approaches infinity off F prime divided by G crime. We're gonna get one plus x co sign of X plus the sign off acts old abided by two acts dividing the numerator and denominator by acts and then playing infinity. We're gonna get that This is equal to zero plus plus the co sign of infinity plus zero all divided by two. So since cold cider backs oscillating between negative one and one, this limit does not exist. Finally, for part D. So for party, if we can just All right, So for party, we're gonna say that Yes, because by definition of low petals rule, if the limit of F prime divided by G prime exists, then the limit off f divided by G off the limit of F kind. But G time for indeterminant forms such as this case is infinity divided by infinity.

For this problem we are asked to find the limit as X approaches infinity of sine of X squared over X. Using the squeeze the squeeze zero. So the first thing that we can note here is that sine of X squared it's going to be greater than negative one or less than positive one. No matter what value of X we put in. Which then means that one over X is going to be less than or equal to sine of X squared over X, Which in turn will be less than or equal to positive one over X. So what we can do then is applying the squeeze the room. All right, apply S Q T H M So we take the limit as X approaches infinity of negative one over X. You can see that that will go to zero if we take the limit as X approaches negative infinity of one or I'm sorry not negative infinity. It's we're approaching positive infinity. We take the limit as X approaches positive infinity of just one over X. That will equal zero as well. And both of those are um in turn less than or equal to the limit. They'll just write limit with blank. The limit that we're concerned about for the first and it's greater than or equal to the limit that we're concerned about for our first or for our function. So putting those together with the squeeze theorem. It's all right. Therefore using the shorthand by sq th em by squeeze them we have that the limit as X approaches infinity of sine of X squared Over X Must Equal zero.

In this question we have to find the limit as X approaches minus two. For the function access where plus two X plus one. I knew I'd buy access word plus X plus one. So first of all we have to substitute here -2 So we get -2 is worth. Let's do in 2 -2. Last one Only. Weird Way -2 is what? Less -2 and less one. We can see that in the stands out and took four minus four plus one. Divided by 4 -2. Last one. Okay, we can say that This can be written as one x 3 and you can see that it is not intermediate form. Okay, not indeterminate from so for this reason we can say that the answer of this question is absolutely one by 33 years. Okay because there is no need to use here. I lost spittle because there is not intermediate form here. Okay means there is not zero by zero Or infinity by can you to form? So the answer is one x 3. Thank you.

So here it's fairly obvious that we're going to need to use loopholes rule because if we plug in zero we're going to get 0/0. So now let's a pilot to Israel will get 1/1 plus route X times um one half X to the negative one half. So that's going to be the first portion and the bottom is going to be um the bottom here is going to be one half facts to the negative half. Based on this, we see that this can now go on top Because excellent, negative one half. That'll make it to the positive 1/2 and that will cancel out with this. So when we evaluate this we have We can eventually evaluate the limit as X approaches zero from the right.


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