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NoteenaeMAA Heuswownner Cuc #CUVT. C0jLV2 S0. _ 19 Lont;pset6: Problem#ine Lrialb Iype DNE5'(3)Khua FnetettBWorkPaobtrd8>517ttDean OaniHeltana Nt ett "...

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NoteenaeMAA Heuswownner Cuc #CUVT. C0jLV2 S0. _ 19 Lont;pset6: Problem#ine Lrialb Iype DNE5'(3)Khua FnetettBWorkPaobtrd8>517ttDean OaniHeltana Nt ett "'I"a Etoent J5 G0-2 #_Cnf 4

Noteenae MAA Heuswownner Cuc #CUVT. C0jLV2 S0. _ 19 Lont; pset6: Problem #ine Lr ialb Iype DNE 5'(3) Khua Fnetett BWork Paobt rd 8>5 17tt Dean Oani Heltana Nt ett "'I"a Etoent J5 G0-2 #_Cnf 4



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Ilcrc, whire precipirare $\{\mathbb{R})$ obraincd on rrcarment wirh aqucous solurion of $\mathrm{BaCl}_{2}$ is of (a) $\mathrm{AgC}$. (b) $\mathrm{CaCl}_{2}$ (c) $\mathrm{BaSO}_{4}$ (d) $\mathrm{PbC}^{\prime} \mathrm{I}_{2}$

So at this limit we can see that if we try to directly substitute in for for Z, we're going to get zero in the denominator and that is not allowed. So you can't directly substitute in for here. And there's not really a way or an easy way at least to try and reformat or try and reduce this um expression somehow so that we can get something else in the denominator. So what I'm gonna do instead is be looking at the right and left hand limits of our function and if those two limits are equal to each other then we will have a limit here limit will exist and will be equal to whatever that value is. So let's look at this um Right Hand limit 1st. So the right hand limit that means we're coming from values that are greater than four. So if you're on an X. Axis were coming from the right of four. And that means that since we have values greater than four for Z, if we minus four from z, that's always gonna be positive since C is greater than four and then if we raise it to the sixth power that's still gonna be positive. And Z squared plus one when Z is for is also always gonna be positive. So here are numerator is going towards Four squared plus one, so 17 and our denominator is going towards zero and it's coming from a positive side. So that means that this is going to be equal to positive infinity. Let's go ahead and do the same thing with this left hand limit. So we're coming from values that are left less than four. Or if we are on the X axis were coming from the left of four. And so that means that Z minus force and see is less than four is gonna be negative. However then once we actually raise it to the six power, that's going to become a positive value. So we're going to have a positive value in the denominator and our numerator is still gonna be positive. So this left handed limit is also going to be equal to positive infinity, which means that the double sided limit as he goes to four, is equal to infinity.

To do this limit, we're first going to cut it up into three different limits since it's the limit of three terms or the sum of three terms, which is equal to the limit of the first term just to see cubed plus the limit of the second term, just three Z squared. And then lastly plus the limit of the last term, Just just five, which is a constant. And whenever we have a limit of a constant, that is going to be equal to that constant, no matter what value of Z we're looking at here. So this limit here is going to be equal five. Um what we can do to these other two limits is take out each of these constants. Um so we can say this is equal to two times to limit, Zika is negative, two of c cubed plus three Times. To limit as he goes to negative two of c squared. And then we already saw that, that limits equal to five. And whenever we have a limit of a function that is just raised to some power. In this case we just have Z raised to the third power and the race the second power. Um that's going to be equal to the limit of that function. So in this case the function of BC then cubed. So we had Z cubed and we were looking at the limit of Z cubed. Now we're looking at the limit of Z and then cubing that limit and then we can do the same thing here and have the limit as he goes negative too of C squared plus five. So these limits are both going to be equal to -2. So this is going to be too times negative two cubed plus three times negative two squared plus five -2 cubed is equal to um -16. So or sorry negative eight. And then we have times two would be negative 16, three times 4 is 12 and plus five. So 12 plus five is 17 -16. It's going to be equal to one.

In this problem we are going to do composite functions with a function named F. Of X. And a function named joe backs. So F composite G. Fx me ask him deposit. G means F. Of G. Of X. Which means take your F. Functioned and where there's an X. You're going to put G. Fx in there. So this would simplify to be six X plus three. There would be a linear function and the domain would be all real numbers. Next G composite F means take your G. Function means she of F. Of X. Which means take your G. Function. And where the X. Was. You're going to put your F function in there. When we simplify that we get six X plus nine. A linear equation. The domain is all real numbers. F composite F means F of F. Of X. Which means take your F. Function. And where there was an ex put your F function in there again. This simplifies to be four X plus six plus three or four, or Y equals, sorry, Y equals Y equals four X plus nine. And this is a linear equation with the domain of all real numbers. Then G composite G means G of G of X, which means take your G function and where there's an axe, put your G function in there again. So this would be um y equals nine X, which is a linear equation and the domain is all real numbers.

So we're going to find this limit using the properties of limits. However first what I'm gonna do is make this a lot easier on yourself or on myself um by factoring what's in this limit. So what we're looking at And what I'm gonna do is take out a -2 in the denominator. So we have negative too, multiplied by 3 -2 y. And we can see here that when we distribute this negative two we get the negative six and we get the positive for Y. And then on the bottom we also have this three minus this too. Why? And we can just cancel these two terms out. And so this is going to be equal to the limit. That is why it goes to negative three of just negative two. And we know that the limit um of a constant is always going to be equal to that constant. So in this case this is going to be equal to negative two.


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