anonymous
  • anonymous
Can someone please explain the solution of the following limit to me? I'd really appreciate it!
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
AravindG
  • AravindG
Well first to check if the limit exist we plug in -1
AravindG
  • AravindG
Here 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.

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anonymous
  • anonymous
Right, I understand everything until it brings out a 3/2 from the limit
AravindG
  • AravindG
You can see that once cancellation is done after factorizing the limits tends to a value.
AravindG
  • AravindG
3/2? I am getting 1/4
AravindG
  • AravindG
\[\dfrac{1}{(-1)(2)(-2)}\]
anonymous
  • anonymous
Yeah i mean here... \[\frac{ 3 }{ 2 } \lim _{x \rightarrow1}\frac{ x-1 }{ \left| x-1 \right| }\]
anonymous
  • anonymous
where does the 3/2 come from?
AravindG
  • AravindG
Is this the same question?
anonymous
  • anonymous
yeah, check the solution in the attachment
anonymous
  • anonymous
Ohh my bad wrong attachment haha wait a sec
AravindG
  • AravindG
I dont see 3/2
AravindG
  • AravindG
:|
anonymous
  • anonymous
anonymous
  • anonymous
in this one XD sorry lol
AravindG
  • AravindG
(x+2)/(x+1)=(1+2)/(1+1)=3/2
anonymous
  • anonymous
so why isn't x plugged in for the rest of the equation?
AravindG
  • AravindG
They split the limit as lim xto1 (x+2)/(x+1) times lim xto1 x-1/|x-1|
AravindG
  • AravindG
lim(g(x)f(x))=lim(g(x)) times lim(f(x)) Thats what they used
AravindG
  • AravindG
Get it?
anonymous
  • anonymous
Ohhh yeah now i get it
AravindG
  • AravindG
well done :)
anonymous
  • anonymous
but why? since there's an absolute value?
AravindG
  • AravindG
You 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
  • AravindG
Yeah absolute value means we need to observe the function under regions.
anonymous
  • anonymous
can't i just cancel out the two (x-1) ?
AravindG
  • AravindG
Nope.
anonymous
  • anonymous
and just get 3/2
AravindG
  • AravindG
|x-1|=x-1 when (x-1) >0 |x-1|=-(x-1) when (x-1)<0 This comes from definition of absolute value function. |x|=x when x>0 |x|=-x when x<0
AravindG
  • AravindG
So the limit f(x) shows two values in 2 regions.
anonymous
  • anonymous
So 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
  • AravindG
Limit exists for absolute value function. The only magic is that it exhibits different limits in different regions.
AravindG
  • AravindG
That is why when we have absolute value function we give the answer in regions.
anonymous
  • anonymous
Mhmm i understand...
anonymous
  • anonymous
Im just thinking that I wouldn't have initially thought of approaching the problem like that (splitting it to 2 limits)
AravindG
  • AravindG
Notice 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
  • AravindG
Properties like these are used by those who approach the problem rationally. It makes it more elegant.
anonymous
  • anonymous
Oh okay, that makes sense
anonymous
  • anonymous
you 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
  • anonymous
That would eventually give me the same answer right?
AravindG
  • AravindG
Of 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
  • AravindG
You could do all these cases separately and evaluate it in your notebook to make it 100% clear.
AravindG
  • AravindG
That way you get more than one way to approach a problem.
anonymous
  • anonymous
Awesome! :D Thanks so much @AravindG !!! I bugged you a lot haha I'm sorry! :D
AravindG
  • AravindG
haha I like being bugged. It means the learner is a good student :)
anonymous
  • anonymous
Lol Thanks!! :D i appreciate it!
AravindG
  • AravindG
yw ;)

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