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Lot g(r) = r'_2x+7 . Evaluate oach d the ddlosial Hnlts (a) Jimg6*) =(b) lim g() =(c) lim g(r)...

Question

Lot g(r) = r'_2x+7 . Evaluate oach d the ddlosial Hnlts (a) Jimg6*) =(b) lim g() =(c) lim g(r)

Lot g(r) = r'_2x+7 . Evaluate oach d the ddlosial Hnlts (a) Jimg6*) = (b) lim g() = (c) lim g(r)



Answers

Suppose \lim _{x \rightarrow b} f(x)=7$ and $\lim _{x \rightarrow b} g(x)=-3 . Find
$$\begin{array}{ll}{\text { a. } \lim _{x \rightarrow b}(f(x)+g(x))} & {\text { b. } \lim _{x \rightarrow b} f(x) \cdot g(x)} \\ {\text { c. } \lim _{x \rightarrow b} 4 g(x)} & {\text { d. } \lim _{x \rightarrow b} f(x) / g(x)}\end{array}$$

The following problem. We want to Suppose that the limit as X approaches be of FX is seven. Um the limit as X approaches be of G of X is -3. So we want to find the limit as X approaches be of F of X plus G of X. And we know that we can use our property. That we have to say that this is equal to the limit as X approaches be uh Mathematics plus the limit as X approaches be of G of X. And we know that ffxf 27 G of X is negative three. So we'll end up getting four. Then the next one. Since we're multiplying, we can use a limit property where we say the limit as X approaches, we have ffx, I am the limit as it approaches the of G of X. Again we have seven and negative three. So it's going to end up giving us a negative 21. Now we want to the limit a four G. X. So that's gonna be four times the limit of G of X. So that is going to end up giving us um -12. Their final answer

So you know Oh, it's no limit off. Good begs us X approaches. See people too. If you're thinking minus your beauty miners, they're important on the limit. Well, thanks Off Geo picks a six support. You see, she won't go see No, uh, that computes first a there is a limit. Six approaches. See off this home if plus g uh, so again he's give me Well, these new function, this song is gonna be You know, it's all the functions Is gonna be a new potion you can't bury getting dude on some of our computer Believe it's separate. Does he go to leave? It affects us except with dizzy plus the limit. Oh, your fix Zixia purchase. Okay, you the d approaches bridge, ISI Uh so here we have all minus infinity plus three, Honestly plus three That is gonna go to minus. So he's wrong being No, Uh, what if you have the product? There is a limit. Sex approaches. See off. You will fix Joe picks? Yeah, It's, uh we can do the same begun. Well, obviously you can't bring it into the product of the two different thing. It's since those separately excused. It's gonna be leaving us Air exit. Butch is see if they lead us. But you see six about to see off. So we're gonna My disability. That's three. But that people too. So, uh, well, now with the same conscious what we wanted to compete believing it. Six. Abort you see off Gene, He did it by half. Uh, so? Well, since, uh, Gee exists weaken. You can't break it into they live it off. Six of what? You see. Six g exists. You come bring it into that. Leave it or leave it. Oh, because of these, always that. Leave it here on bottle. Means knows you. But it wasn't serious. You asshole. You have three over, baby that. Who is this you, you know, by by a very large number, Good spirits. So, in living these zeros off this raid, both of you

Knowing that the limit as X approaches be of F of X is seven and the LTD's X approaches be of G of X is negative. Three. We could take each of these top limits independently and then add the results. And so that would end up giving us a value of seven for F plus. The limit of G as expert's be would give us plus and negative three. And so those would combine to four so we can always split up to limits. If they're added or subtracted like that, the next case. We could take the limit of each of these and then multiply the results together. So very similar concept the previous one take the limit of F won't apply that by the limit of G, and then that would end up giving us negative 21 for the second the evaluation of the second limit, the next one a constant. We could move up front and then multiply it. By the limit of G, the limit is expertise be of G of X, which is negative three. So that will get us negative. 12th and then finally we could take the limit for the last problem of this rational expression, the limit of the top divided by the limit of the bottom. And that would end up giving us seven divided by G, which is negative three. And so we could just leave that a seven over negative three, or move that negative up top or just move it out in front. And so negative 7/3 is what that final limit evaluates to.

Okay, so we're giving the falling limit and their values and were asked to find the fallen. So for court A, I'm last to find the limit as X approaches be of the fallen. So we see that we can start by using our some role plus role so we can rewrite that as the limits. As X approaches be, uh, f of X, plus the limits as X approaches be of Kiev X. Now, that's equal to the limit of F seven, and the limited G is negative three. So we get seven minus three, which is equal to four now, for part thes, we have the limits as X approaches be of f of x times. Do you have ax? So in this case, we can use our product herbal. So we have the limits as X approaches be of FX times the limits as expert to be of G of X. So we said the limit of after seven and the limited G is negative three. So we get negative 21 the report seeing we have following limits. Okay, so we see that we have a constant in front of G, so we can use our constant multiple full and pull out our constant Do we have four limits as X approaches be of G of X. Well, that's equal to negative three. So we have four times negative three, which is equal to negative 12. And last week for 40 we have the falling, so we can start by using our quotient room so we can rewrite that as the falling. Okay, so we know older limit of that seven and a limited G is negative three.


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