5

2 49x"15x 1149x2 5a + 20lim T 0...

Question

2 49x"15x 1149x2 5a + 20lim T 0

2 49x" 15x 11 49x2 5a + 20 lim T 0



Answers

$5-28$ Find the limit.
$$\lim _{t \rightarrow \infty} \frac{\sqrt{t}+t^{2}}{2 t-t^{2}}$$

Okay, So for this problem, we're asked to look at the long term behavior as he goes to infinity of this function of teeth. So we asked, to which value this is converge if it converges. And let's look at this so called back from section 2.1 that we were given some really helpful rules regarding the sums and cautions of sequences where we had a limit of it. Some of sequences was equal to the sum of the limits of each sequence, and we had that the question of sequences was equal to, um, the ah, the quotient of limits. And to be more precise, let's recall that this, of course, only applies if we're not trying to divide by zero. So we have to make sure be a end is not converged to zero first before we employ this Now and this problem, we are obviously trying to simplify this expressionist complicated expression of tea, but it's not quite obvious how to do that, And it might be a first a good idea to try to bring these limits ends that we're talking about. The limit SC goes to infinity of T minus the limit as he goes to the minus the limit as he goes infinity of t's times the square root of tea over et cetera. But if we try to do that, then right off the bat, we get an infinity minus infinity over infinity, plus infinity minus five. And so we may have jumped the gun a little bit on that. And so this the book actually offers a really, really, really helpful technique where we take an expression like this and we divide by the highest power in the denominator. So we'll divide each term by t ah to the three house. And the reason is works, of course, is that all we're doing is multiplying this quotient by something of the form A over a, which is always equal to one. And that over in this case, is a, um is equal to one over a t 23 halves over one over a t to the three house. And that's another way of writing won over t to the three house times Teacher three house over one, which is, of course, a over a which is equal to one. So if we do that, we continue on here that we have the limit of T over T to the three halves minus t time to square of tea over teeth, three halves asked. He goes to infinity over, uh to Mrs. Cancel the 2 to 3 halves over 2 to 3 halves right off the bat, plus three t over a team to the three halves, minus five over and teach the three house. Now if we were, call some girls from exponents. If we multiply to powers, if we multiply t raise some expert and then we add the exponents. We had the powers. So if we have t to the A times t to the B, then it's t to the A plus B and similarly, for quotients. It's the opposite that so we subtract the powers. So in the first term will be subtracting t over. We subtracting. Ah, tea from we'll be starting absolutely. Subtracting three teacher three halves from t But T is equal to t to the two over to so we can simplify that really easily and right Ah one over. Teach the 1/2 because they're subtracting ah to over to minus three over two, which is minus one. So we have to to the 1/2 uh, inverse of tea to 1/2 minus and t times. The square root of tea is equal to tee times to over two t to the two over two times T to the 1/2. So if we add those, then we have tea to the three house. Then we have tea to the three halves over teach of the three halves, which is equal to one. So this is equal to one. This is equal to ah t to the minus 1/2 which is equal to won over t to the half or t one over T Thio squared up to you and on the bottom, we have to plus and similarly, here we have three over t to the 1/2 minus five over. That's a really ugly five five over a t to the three halves. The limit is he goes and Vinny. Okay, now I know I'm thinking looking pretty good here. We have much more tangible problems. Especially want to consider another rule from 2.1 where if we have a limit of a sequence and and 2.1, we were only talking about integers um, if we're talking about in like this. But of course, the functions of functions of, ah sequence of ah t said can be taken any value. Um, but at any rate, um, definitely difference between 2.12 point two. But as engrossed infinity, we had won over and to the P is equal to zero always if P is greater than zero. So what this is saying is that we have to have the inverse of some number raises, some power then that is always gonna convert zero because the denominator will blow up, he'll become increasing. And the term is a whole one over. A bigger and bigger number is closer to zero. And so, if we if we employ that here we see that once we bring these limit signs in, we're gonna have limits of one over t to some power that's greater than zero. And so when we write the limit of see going to infinity Ah, one over t to the 1/2 or, you know, Tito the minus 1/2 or one over the square root of tea. However you want to call it, uh, that's gonna converge to zero now the limit of one nasty ghost Infinity is is one. So it's it's constant and we have to on the bottom, plus again the limit as he goes Infinity of three over t to the 1/2 uh, three times T to the minus 1/2. That's gonna converse to zero for the same role here and minus limit a sea ghost Infinity of five over t to the three Have it's that always goes here, huh? Death. It'll screw zero. And so now this whole term that we have left is minus one over two. Well,

We want to find the limit as X goes to infinity of two expert plus one all squared, divided by ex mice one squared times X, where plus x bursting re probably want to check is to see if we could just evaluate this directly. So now I'm gonna put quotes around this because we technically can't do this. So let's see well, the limit as to extra plus one as X goes to infinity. Well, that's just a proble facing upward. So that should go to infinity. And if we square something going to infinity, that should still go to infinity Now in the denominator x minus one is a line and it would be going towards infinity for this one since our slope is positive squaring that should still give us a positive ID me So we get positivity times and then x squared plus X. That's also an upward facing problems that goes to infinity. Now I have absolutely no idea what this means because well, first infinity times, infinity would still be something going towards infinity. But I still don't know what this expression means. Like what does it mean to divide infinity by infinity? So What that tells us is we need to do a little bit of algebra to put this into something that we do know. And the trick that we learned was to find the overall largest power, the numerator or denominator, and then multiply by that reciprocal. So in the numerator, well, we have that to x squared plus one. But if we were to boil this out in the numerator, we really have and X to the fourth power. Not in the denominator. If we were to oil out X minus one square, that gives us X squared. And then we have a X squared right next to it. So the denominators are just power would also be X to the forth. So what we're going to do is multiplied by the reciprocal of next to the fort in the new Moran denominator. And doing that, we can rewrite it in the following way. So first in the new mayor, I'm going to rewrite one over X to the fourth as one over X squared squared, and it will make a little bit more sense. Why would I do that in a moment? And then I'm gonna rewrite one over x to the 4th 1st as one over X square than the X minus one squared that we have X squared plus X over here. And then I'll just have one over x squared. So this year was the one of her ex to the fourth term. Now, the reason why I want to rewrite it like this is we can do a little bit of Alvin, bro. Well, one of the rules for exponents is if you have something that same explanation, you just pull it out. So this here would become two x squared, plus one times one over X squared, which would give us not ex but two plus one over X squared squared. Then this one here, we could do the same thing. Like we'll be dead in the numerator. Pull that square out, and that should give us one minus one over X square. And then, lastly, this one right here well, we were distributed across would be one plus one over X. But I almost forgot the limit. Then we still have our limit as X approaches infinity out here. Now, If we were to applied the limit now, well, all of the ones words like one over X. All those will go to zero because it be one of her infinity, which is the same thing, is asking us what is element as we keep dividing by something larger and larger and those would all go to zero. So these go to zero so we can rewrite this US two plus zero squared all over, one minus one silver it more to look one minus zero squared and then one plus zero. So we can simplify this town to be two plus zero squared, which is going to be just two squared or or in the numerator and then the dominator one my service one square that get one and then one plus zero is one so that one times one is one. So the limit as experts affinity of this rational function should be for

Today we're going to solve problem numbers. If here, let me do you tend to minus three The square minus nine divided by to the square plus 70 plus three. We have to check whether they limit access or not. So we can write like limit. He tends to minus three The ministry enter T plus three divided by two D plus one en duty plus three limit Didn't still minus three. So do you Ministry? The ministry can be canceled. We can write like t minus three divided by to keep us one. So limit Do you tend still ministry the ministry? And if I didn't buy limit data and still minus tree to take this one So we will get like substitute equal to minus three, minus three minus three. Rewarded by two in the minus three plus one minus six. Developed by whiteness Fight which is equal to six by fighting So six by five limit exists. Thank you

We want to find the limit, Stu Part infinity of e race to the negative one over t squared. Now if we were to just plug in infinity directly and I say plug and in quotes, this would give us so it b e raised the negative. So we have well, t Square would go to infinity and then one over, so we'd have won over infinity like this. So if we keep dividing one by something that's getting infinitely large, that should go towards zero. So this would really be e race to the negative zero, which is just eat zero or what? So the limit of this should be one.


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