5

H puu [Eule" = 1 1 U W Ul 454 Linit entampicrapplicatian Involvina yeclor CLculu: 1 1 Wed uld" 1 1 ] 1 It uDJJ 1 1 1 WLCUCA LInW hlcae1...

Question

H puu [Eule" = 1 1 U W Ul 454 Linit entampicrapplicatian Involvina yeclor CLculu: 1 1 Wed uld" 1 1 ] 1 It uDJJ 1 1 1 WLCUCA LInW hlcae1

H puu [Eule" = 1 1 U W Ul 454 Linit entampicrapplicatian Involvina yeclor CLculu: 1 1 Wed uld" 1 1 ] 1 It uDJJ 1 1 1 WLCUCA LInW hlcae 1



Answers

$$A=\frac{1}{2} h\left(b_{1}+b_{2}\right), \text { for } h$$

In the problem we have 56. And this is yeah it is gen X get works. Get three X. Get your necks. It's two weeks is tricks plus if one X after works three X gen X G networks. GTX H one X. HdX history X plus. If you're next afterwards after the X. And the one x. Networks G three X H one X H two x. and three x. No further. We have fds a. And this equals two. If one yeah. Have to wear after a year. Do you want to G two A. Gayatri? H one A. It's too good. Is three. Okay. And these are reputed To two times more studies if one A. Have to A F three given a and geeta, Geeta A 20 H two A. It was today glass if one after where have today do you want to? They were the tree H one A H two A is three. Now this equals to zero plus zero plus zero which is equal to zero. Since F R A Is equal to zero. G R A is equal to H R A. Is I mean this equals to 04. All the times that is if R0 gr hr is also zero For Article 1, 2 and three. So we have this as the answer to the problem

So we have to evaluate the following limit and see if it equals one over E. So um the first thing to do is to plug in the value of the limit and see if it's something easy. So if I plug in the value H equals zero. I get one minus zero raised to the negative 1/0. And since you can't divide by zero, this becomes a problem. Now in order to solve this problem, we need to use one rule called low petals rules. So I just want to remind everyone what low petals rule is. So let's suppose that you have a function actually a quotation of two functions and you're taking the limit as X approaches some about you And you can write the expression as a ratio of two functions. Now if this evaluates to an indeterminate form that either looks like 0/0 or infinity over infinity and this is plus or minus infinity. Then you can rewrite the limit as follows. You can take the limit of the derivative of the numerator divided by the derivative of the denominator. So I just want to quickly interject and say that this is not the corset rule. Look up the quotient rule, remind yourself that is nothing that has nothing to do with this. This just says that you isolate F and take the derivative of that and then you take the derivative of G. And then you compare the slopes. Okay? So it essentially gives you an easier way of evaluating the limit because usually the derivative is a slightly simpler function, especially with polynomial. So that being said, let's evaluate this limit. The first thing I'm gonna do is I'm actually going to set the limit equal to evaluate why. You'll see why in just a second. So the limit h approaches zero of one minus H. To the negative one over H. I'm going to let that equal why. And the reason is this exponent is quite pesky in order to get rid of the expo and I'm actually gonna take the natural log of both sides. Now you can interchange a function and a limit as long as the function is continuous and in this case the natural log is continuous. So I'm gonna go ahead and say that the natural log of Y equals the limit as a jew approaches zero of the natural log of one minus H. To the negative one over H. Now, because the properties of logarithms, I can move this exponent down as a coefficient and then I get that the natural log of y is equal to the limit. As a jew approaches zero of negative one over H times the natural log one minus H. Just to remind you, we are trying to solve for why why is equal to the limit that we want? This is why. And here's the limit that we want in blue. Now to evaluate this limit. Um I can rewrite this as the limit As a church approaches zero of the Natural Log. Well negative the natural log of 1 -3. All over H. Okay. And if I plug in H equals zero at this point so I go and just plug that in, I will get negative natural log of 1 0 over zero. Well the natural log of 1 0 is the natural log of one And the natural log of one is 0. So I was able to write a ratio of two functions. And upon evaluating the limit, I get an indeterminate form. So now I can use low P. Tall and lo petals rule says that the same limit which was equal to the natural log of Y. Is the same limit. But now I separately take the derivative of the top and the bottom. So I'm actually just going to write that out. Just we don't confuse it with the caution rule. And I should specify that I actually made a mistake. Um It's not a big mistake here. We're taking the derivative with respect to H. I never use H. That's kind of line thrown up. I always say as X approaches something. So regardless, anytime you take a derivative, uh this variable in this variable better match. So that was my bad. Okay so this is just gonna be H. Now the derivative with respect to H. Of negative natural log of whatever. I can factor the negative out. And the derivative of the natural log is just one over whatever's inside. But I have to remember to multiply by the inside function using the chain rule. So I also have to take the derivative of um one minus H. And that's just negative one. So these negatives cancel and the derivative of um H. Is just one. So I have to remember that I can take the same limit but now I evaluated those derivatives. Okay, so just to rewrite what we've done the natural log of why we use loopholes rule in the numerator and denominator. And we said that now I'm going to take the limit, I'll be slightly simpler function. So these negative signs go away and I'm left with 1/1 -3 divided by one. Well that limit is super easy because now I can directly substitute the value at H equals zero And I get 1/1 0 which is one. I'm not done. The limit does not equal one. Because the limit that I'm interested in is why I set the limit equal to Y in the very beginning. So the natural log of Y is equal to one. So if I exponentially both sides, I get that. Either the natural log of Y cancels to why and the limit that I'm interested in why is equal to E. To the first power, which is just so this is the limit and it's actually not equal to um one over Eats. This is false.

We want to take 1/6 G to the fourth age to the fifth times 36 g to the seventh ht 11th. These are still mono meals. We just They're gonna have ah, a few extra steps than maybe we have in some of the earlier problems. Our goal is still the same. We want to multiply the things that go together, meaning we want to multiply the two coefficients together. This time we have two different variables. We had the G variable and the H variable. So in separate steps we will multiply the geese together and then another step. We will multiply th together. We'll start with the coefficients. So are two coefficients are 1/6 which we want to then multiply to other coefficient, which is 36 16 times 36. That's what we're gonna be doing. We also have G variables. So we would take G to the fourth times, the other G variable of G to the seventh. We also have our H variables, which means we would want to take aged to the fifth time's the other H term of age to the 11th apps on its route that black instead of red. Sorry. Here we go. 1/6 times 36 is essentially the same thing as just taking 36 divided by six. In either case, you get six G to the fourth times G to the seventh. Means you're really taking four plus seven to give you G to the 11th H to the fifth times h the 11th also would mean you're adding these exponents together. So five plus 11 would give us aged to the 16th, making our final answer six g to the 11th Power H to the 16th power. That would be your final answer.

In this problem, this compound, which I am writing here, just look at it carefully here. It will be A and here it'd pr and this compound here it is. Beyond here it will be a These two compounds in Potenza access any manage to then as to oh, work up https your four As two years of four and me oh, edge to produce the final, productive this compound reality. They will won't know so according to the option in this problem absence. See absentee are correct answer.


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