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anonymous
 one year ago
Can someone please explain the solution of the following limit to me? I'd really appreciate it!
anonymous
 one year ago
Can someone please explain the solution of the following limit to me? I'd really appreciate it!

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AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Well first to check if the limit exist we plug in 1

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Here we get 0/0 form so we try to factorize the denominator and see if any cancellation is possible with numerator. In this case while factoring we got an (X+1) in denominator that we could cancel out in numerator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, I understand everything until it brings out a 3/2 from the limit

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2You can see that once cancellation is done after factorizing the limits tends to a value.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.23/2? I am getting 1/4

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2\[\dfrac{1}{(1)(2)(2)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah i mean here... \[\frac{ 3 }{ 2 } \lim _{x \rightarrow1}\frac{ x1 }{ \left x1 \right }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where does the 3/2 come from?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Is this the same question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, check the solution in the attachment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh my bad wrong attachment haha wait a sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in this one XD sorry lol

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2(x+2)/(x+1)=(1+2)/(1+1)=3/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so why isn't x plugged in for the rest of the equation?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2They split the limit as lim xto1 (x+2)/(x+1) times lim xto1 x1/x1

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2lim(g(x)f(x))=lim(g(x)) times lim(f(x)) Thats what they used

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh yeah now i get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but why? since there's an absolute value?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2You see g(x)=lim xto1 (x+2)/(x+1) is a limit that exists but for f(x) we needed more work so we took out known limit outside and dealt with f(x).

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Yeah absolute value means we need to observe the function under regions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can't i just cancel out the two (x1) ?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2x1=x1 when (x1) >0 x1=(x1) when (x1)<0 This comes from definition of absolute value function. x=x when x>0 x=x when x<0

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2So the limit f(x) shows two values in 2 regions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what if we find 3/2 and say that since its an absolute value function, its either 3/2 or 3/2 so the limit does not exist

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Limit exists for absolute value function. The only magic is that it exhibits different limits in different regions.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2That is why when we have absolute value function we give the answer in regions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Mhmm i understand...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im just thinking that I wouldn't have initially thought of approaching the problem like that (splitting it to 2 limits)

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Notice that you could do the problem without splitting also. We do the same operation on f(x) with the g(x) part remaining a person with no specific role. So to make it more appealing we split the limit.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Properties like these are used by those who approach the problem rationally. It makes it more elegant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, that makes sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you see at the end it breaks the lim into one with x > 1+ and x>1 ? Would it be correct if i do that from the very start and compute those 2 limits separately?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would eventually give me the same answer right?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2Of course it would be right. The limit should be answered in intervals and it is in the hands of the student to decide when to split it to intervals either in the end or during the start.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2You could do all these cases separately and evaluate it in your notebook to make it 100% clear.

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2That way you get more than one way to approach a problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! :D Thanks so much @AravindG !!! I bugged you a lot haha I'm sorry! :D

AravindG
 one year ago
Best ResponseYou've already chosen the best response.2haha I like being bugged. It means the learner is a good student :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol Thanks!! :D i appreciate it!
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