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Given that

$ \displaystyle \lim_{x \to 2}f(x) = 4 $ $ \displaystyle \lim_{x \to 2}g(x) = -2 $ $ \displaystyle \lim_{x \to 2}h(x) = 0 $

find the limits that exist. If the limit does not exist, explain why.

(a) $ \displaystyle \lim_{x \to 2}[f(x) + 5g(x)] $

(b) $ \displaystyle \lim_{x \to 2}[g(x)]^3 $

(c) $ \displaystyle \lim_{x \to 2}\sqrt{f(x)} $

(d) $ \displaystyle \lim_{x \to 2}\frac{3f(x)}{g(x)} $

(e) $ \displaystyle \lim_{x \to 2}\frac{g(x)}{h(x)} $

(f) $ \displaystyle \lim_{x \to 2}\frac{g(x)h(x)}{f(x)} $

a. -6

b.-8

c. 2

d. -6

e. $\lim _{x \rightarrow 2} \frac{g(x)}{h(x)}=$ Does Not Exist (DNE)

f. 0

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in this problem, it is given that limit, fx limit extends to two of fx equals two fold. Right, limit extends to two of fx G x equals two minus studio. Right. And limit hx and limit is given extends to two is equal to zero here. Right? So we need to calculate whether the limits exist or not and we need to explain why. Right, So this is very simple. We need to put all these values indeed all parts, then we'll see whether the value exists or not. So let us see how do we do. So, this is given limit extends to do of you can see the whole bracket is uh under that bracket, F X plus F five G x is given to us. So what it is so can be right here limit extends to two separately for a fix As well as a plus for 5, 5 is taking common as it is constant. So extends to do this will be Gs So we will put the value here, what we will get we know fx extends to to hear it is given So we will just put the value that is four here, can we put Yes. So here, before plus G x is nothing but five into here by five year olds taking comments five in towards but I want to minus two. Right, so this will finally give us minus six. Right, so this is the phone number, so we can see limit exists, leverage exists as Yeah, it contains really. Mhm Right, No. Right, so this is an answer. Right? So this is not whatever comes in the form 0x0 or infinity by infinity or something like that, that something is upon zero or something upon infinity. So that will that answer will be not existing rate. So here we will see again we've been right here for uh second problem that is partly limit extends to two G X hold you so can we right here like this? That limit GX-right extends to to this is all as Q right. So we can put the value, we know that this value we are given with so this is -2 to the power to see that is nothing but -8. So limit exists, limit exists late. Similarly here, we can see we can also write this third equation. We know this can be written as limit Rotor will limit extends to two ethics. Right? So this will be the king but this will be true to work four because effects we know that limit was extended through then the value is for here we are given. So this value is to So limit exists. Thank you any any number we are getting in a whole number or a real number we are getting then the value exists Right? No, for the deep part we'll see here No, how can we solve it? This is given by three limit Extends to two. FX. This is divided by Limit Exchange 2, 2, this is DX we can also try to eat us. We will write given the question that is saying to Given values people fortunately right? So 3-4 upon -2. This will give us -6. So limit exists again. No, for the e part we can see here Epics is given to us that is let us right, we can write it again in this way limit extends to do a bitch That's divided by limit extends to two edges. Okay, so here the value for this is poor And upon zero. So this is going to win politic. Therefore LTD does not exist. Yeah. Okay, Okay, so now for the last and the f part you can say limit extends to this is DS then again, this is multiplied with limit extends to do of a chicks right? This is divided by limit extends to two of fx. This will be given by work, this will be for G Exit is -2 values and for a checks it is zero and this is her upon for so this is zero. So limit exists, right? So limit exists. So this is how we solve this problem. I hope you understood the concept. Thank you for watching

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