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In Exercises $41-44,$ use the information to evaluate the limits. $\lim _{x \rightarrow c} f(x)=3$ $\lim _{x \rightarrow c} g(x)=2$ (a) $\lim _{x \rightarrow c}[5 g(x)]$ (b) $\lim _{x \rightarrow c}[f(x)+g(x)]$ (c) $\lim _{x \rightarrow c}[f(x) g(x)]$ (d) $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}$

So this is problem 46 were asked to determine the limit as ex A purchase e of a couple, um, different things. So we'll go ahead and start with part a sow with part a were exposed to subtract or add the two. Excuse me. And since our, uh, we'll do limit as X approaches see, uh, five minus two square, this is gonna be actually equal to nine. Now, be well, go ahead and show that so this is gonna be the limit as X approaches. See, uh, six times, five times negative, too. This is gonna be equal to negative 60 and then for part C, it's going to be the limit as X approaches, see of five times G of X over four times f of X, uh, which is gonna be equal to, uh, the limit as ex purchase sea of five times negative two over four times five, which is going to be equal to negative 1/2 on, then finally, for part D. Um, it's going to be the limit. As acts approaches. See, uh, of one over the square root of f of X was just five. It's gonna be equal to one over square root of five. So these are answers right here.

Okay, So this question we're finding the limit as one approaches infinity for a midpoint room on some with and sub intervals on the function that were given is Kovacs vehicle. The 12 minus four x on the interval that we're working with is from 2 to 6 s o. The first step is to write out what the formula is for a midpoint remind some with ends up intervals just kind of just grew out what we're gonna be plugging into So men point reruns on with inseventy rules is equal to Delta Axe times. The sum from J is equal to zero too, and minus one of F uh x sub j plus X sub j plus one over, too. And just as a reminder of Delta X is equal to be minus a over end on B and I are the limits of the interval of an ex subject is equal to a plus J. Delta X and therefore X subject was one just plug in J plus one for Jang right here. Okay, so now we can start figuring out what each of the components is for this problem. So first, let's you Delta X Delta X is going to equal to you might say over end in this case, be a six and A is too so six minus two over end is equal to four over end. So that's where Delta X and our except J is gonna be equal to a, which is to plus J times, Delta X And were you handle two exits four over end. So four J over end Um can actually write X subject as well, So x So Jay pulls one's ari. So rewrite ex subject pulls one, um And that's going to be equal to two plus. So he just distribute this forward to the J please when you get four j plus four over and okay, so now that we have all of our components ready to plug in and figure out what our midpoint months, um, with ends up intervals is going to look like so this is gonna be equal to Delta X. So that means for over end times of some from J is equal to zero. The end minus one. Um, and here we're gonna have s of A except J. So two plus for J over end plus two plus for J plus four overrun. Um, and this is all divided by two. Okay, so now let's simplify this a little bit. So for this inside part here, we just do it all the one step. If you add the tunes, you get four. Um, and these have the same denominators. And just get eight j plus for on. And then we can just divide by two right away on you. Right out. You get four over end times. A sum from J is equal to zero. Toe end minus one from. And here relax. Aft with s of two plus for J over end Closed two over. And okay, so now that we've done this, let me go to the next page and we'll plug that into our function. This is gonna be equal to a four over end times. A son from J is equal to zero and minus one, and our function was 12 minus four x. So we're gonna have 12 minus four times that we had in the parentheses, which is two plus for J over end plus two over. And okay, so now we can distribute this four and subtract from 12 on and When you do that, you get the same well done by the same thing. So I must find my four over and And we have the sum from J is equal to zero and minus one. Um, and here we're gonna be left with four minus 16 J over end, minus eight over end. Okay, so now we can do is we can use the rules of summations and split this up into three parts to just take the sum of each of these components. So that's gonna be equal to four over end times the sum of each. So this same some from J equals zero too. And minus one of four minus on. And we can actually write this coefficient in front, so I'll just put that in front. So 16 over end times, times of some from J is equal to zero toe end minus one of J. Um, And again here we can write it as a coefficient out in front. So minus one over end crimes is some from J is equal to zero Thio end minus one of eight. So now we can again apply our rules of summations. So here, we're gonna be left with four end, uh, times four times and minus one. When you have a constant, you just multiplied by whatever you have on top here. So times so four times and minus one. Um, now we're gonna ply our power some formula. So we have minus 16 over end times and minus one, um, times and over, too. And so I'm just gonna go on to the next part. I ran out of room. Um, and here again, we just have a constant, but we're multiplying by negative one over end. Some of me left with minus eat over end times and minus one. Okay, so now when we simplify this, it's just a little bit of algebra. We can get that this is all equal to negative. 16 minus 16 over end. Plus 32 over and squared. Okay, So this just as a reminder, this is our midpoint, Ramon. Some friends of intervals. So we figured out what that is and the simplest form. And now we're ready to actually take the limit because that's what we're looking for Here were approximating the integral using the summation notation. So let me go to the next page. The limit as an approaches infinity, who were a midpoint. Three months I'm with and some intervals eyes gonna be equal to the limit as a purse. Infinity of the answer that we got on the previous page. So as an approaches, infinity of negative 16 a minus 16 over end, plus 32 over and squared. So now the limit as university new for negative 16 is just negative 16. Um, And here we have an end in the denominator. And when it gets really, really big, this is going to go to zero. Same thing for this right here. So that's just negative. 16 plus zero zero on and I just gives us negative 16. So this is actually our final answer. This is our, uh, limit for this question.

Okay, So for this question were asked to find the limit of the left. We won some with and sub intervals for this function right here on the interval. 123 So first thing to do is to remind ourselves what the formula for the last three months, some with and some intervals looks like. So this is equal to Delta. X times of some from J is equal to zero too, and minus one of F of ex subject. Ah, where Delta Acts is equal to be minus a over end. And being in this case are the limits of the interval, and X sub J is equal to a plus J Delta X. So first thing to do, let's find Delta X. So in this case, Delta X is going to be equal to three, which is B minus one, which is a over end. So it's equal to two over. And so now we can also find X subject on, and that's gonna be equal to one plus to J over end. Okay, so now that we have these two components, we can plug into our formula. So let me want some within seven travels using summation notation for this function would be to over end times. The sum for M J is equal to zero to end minus one of four times one plus to J over end minus issue, is it? Now, if we distribute the premises here, we get to over end times. A sum from J is equal to zero end minus one of four. Minus two is gonna give us too. Plus eight J mover. And now we go to the next page. So that we can do is we can factor out to from our summation. So we do that it's going to get a little bit simpler. We're gonna have four over end times of some from J is equal to zero toe end minus one. I'm gonna for factoring out of two. We're gonna have left for J over and plus one. Okay, so the next step would be to separate this some into two parts sorting the sum of forge over end and the sum of one. So let me write that we still have four over and times, so someone we can just write the foreign on the outside. Since that's a constant for end times of sum from J is equal to zero toe end minus one of J plus. This sum from J is equal to zero to end minus one of one. Okay, so now we can apply our power Some formulas for summations. So we're gonna be left with is for and over. Sorry. Times for over end times. And the formula for just having a J here would be and minus one times and over too. Um, and we have just a constant We just multiply that constant by whatever is on top of the summation. So in this case, that was gonna be and minus one dumb have plus, uh, and minus one. So now this four in this too. It was just a to appear in these two ends. Cancel out. So now we can rewrite this for in times to his aides. We have eight times and minus one over end, plus four times on minus one over. And so in total. That gives us 12 times on minus one over end. Um, and if we multiply this out, we're gonna get 12 minus 12 over. And so this is, um, our expression for the letter months, Um with ends up into roles, all simplified online. Now we have to find the limit of that. So the limit as an approaches infinity of the left re monster some is going to be equal to the limit as an approaches infinity of 12 minus 12 mover and s o the limit of a constant approaching. Anything is just that, Constance. And here we have 12. Um, and this is going to get really, really small as and or the denominator goes higher and higher. So this limit is actually just going to be zero. Some things that are final answer is 12.

So this one really puts everything together? Um, we're gonna take a room in some using ends of intervals and then take the limit as an approaches infinity. So let me start by Just drawing this nine x is pretty steep, depending on how we scale the Y axis, that should do it. And then here's the area we're interested in. It's that triangle. So going from 0 to 2. So this is just a width of two, and this is, ah, height of Well, that point there is to comment 18 So height of 18. So 1/2 base times height area in here should equal 18. So that'll be helpful later. We're doing now is taking a bunch of columns, and the idea is gonna be to calculate that area calculated so generally that we can just say, Well, let the number of columns approach infinity. So let's go to it. The right Raymond. Some has a formula that looks like this, um one to end f of X I Delta X. No, it's putting these formulas. Delta X equals B minus a over end for us right now, that's going to be two minus zero over again. So I'll just say to over end now exit by equals a plus I Delta eggs, which is a plus I times two of her end and I'd rather say to eye instead of times too. So there we go. Um, a equal zero, though. So I want to say like that. So I'm gonna move to another page to do my calculations and, um, I should go. Should go. Well, let's see what happens. Um, we got the limit as an approaches infinity of the sum from I equals one to end. And now the function is nine times x. So when I plug in except I'll have nine times exit by. And here is my expression for that. So nine times I'll write it out for the first time on Thames to ever end. And in times Delta X, which is to over end very typical. You'll end up with II being one degree less than the degree of end on the bottom. So now I've got the limit as an approach infinity of this and nine times two times two is nine times 4 36 I'm gonna just put that all the way out here. Leave the eye here and it's over. End squared right now. But I won't take that end squared and move it past the, um, the Cigna. So that looks like this. So just came from here, over here or there. So anyway, that's convenient way to rewrite this. Now, this whole piece that I have here whips I would have liked that to be a pencil instead of a big eraser. Um, so this whole piece that I have here, you have a formula for this, and I'll remind you of that formula in a moment. It's going to be in times and plus one whole thing over to you. I got one over end squared, right, this in reverse order for some reason. So let's come appear and see what we've got. I noticed that one of the ends would cancel. That's good. And so now I have that will bring this to out as well. So I'll say 36 over, too. I'm so limit of and plus one over end now, this whole thing will approach one the limit. Well, it's easier to see that if you think of it as this. I got an over an and one over end this part one of her end when end gets to be really large, will approach zero, and it's 18 up front to end up with 18 times one. I think of it that way. In other words, 18.


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