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1 Ouold and J8.Ulu O H 1 # li L Tuntintz ch 1 1 11 1 Manul...

Question

1 Ouold and J8.Ulu O H 1 # li L Tuntintz ch 1 1 11 1 Manul

1 Ouold and J8.Ulu O H 1 # li L Tuntintz ch 1 1 1 1 1 Manul



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$\left[ \begin{array}{lll}{1} & {1} & {1} \\ {1} & {2} & {3} \\ {0} & {1} & {1}\end{array}\right]$

In this problem, I can write DDX and Edge. Just look at it carefully. After that, I will write the product name of the compound. So here H g c l two reactive IT four K I. To give the product ID K two, H g A four plus two K c n. This is compounded therefore, according to the given option, option, age, correct and said for this problem, the compound be eat this K two HD iphone. This is the compound. E I hope you understand the reaction of this problem.

And so for the given question, we have a session, given position is Gcl too and SNCf to exist together in equal solution. The reason for this, Given ization is on heating GCS to supply times. So here we have the correct option for the given statement is the option D exertion is falls, but the reason is two. So here we will write the correct assertion for the given reason. So correct position is Gcl too, and SNC and two cannot exists cannot exist together in a glass solution. And the reason is here on heating. Yeah, Gcl too. Sublime songs. So here is false and but are is true. So the correct position is GCL to end. SNC cannot exist together in a solution.

Substituting an age is equal to a be obtained indeterminant form of 00 We can carry out the difference of fractions in in numerator to end up with the limit As H approaches aid off a minus age divided by eight times age multiply by one over H minus. A. Remembering that each one is a is just equal to negative a man's age. We can rewrite this as a minus h two by eight times h well supplied by one over negative a minus h canceling out the common terms. We end up with negative one over eight times each. And if we take a plug in a into H, we end up with negative one over a squared.

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.


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