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

- anonymous

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

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- schrodinger

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- anonymous

##### 1 Attachment

- AravindG

Well first to check if the limit exist we plug in -1

- 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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## More answers

- anonymous

Right, I understand everything until it brings out a 3/2 from the limit

- AravindG

You can see that once cancellation is done after factorizing the limits tends to a value.

- AravindG

3/2? I am getting 1/4

- AravindG

\[\dfrac{1}{(-1)(2)(-2)}\]

- anonymous

Yeah i mean here...
\[\frac{ 3 }{ 2 } \lim _{x \rightarrow1}\frac{ x-1 }{ \left| x-1 \right| }\]

- anonymous

where does the 3/2 come from?

- AravindG

Is this the same question?

- anonymous

yeah, check the solution in the attachment

- anonymous

Ohh my bad wrong attachment haha wait a sec

- AravindG

I dont see 3/2

- AravindG

:|

- anonymous

##### 1 Attachment

- anonymous

in this one XD sorry lol

- AravindG

(x+2)/(x+1)=(1+2)/(1+1)=3/2

- anonymous

so why isn't x plugged in for the rest of the equation?

- AravindG

They split the limit as
lim xto1 (x+2)/(x+1) times lim xto1 x-1/|x-1|

- AravindG

lim(g(x)f(x))=lim(g(x)) times lim(f(x))
Thats what they used

- AravindG

Get it?

- anonymous

Ohhh yeah now i get it

- AravindG

well done :)

- anonymous

but why?
since there's an absolute value?

- 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

Yeah absolute value means we need to observe the function under regions.

- anonymous

can't i just cancel out the two (x-1) ?

- AravindG

Nope.

- anonymous

and just get 3/2

- 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

So the limit f(x) shows two values in 2 regions.

- 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

Limit exists for absolute value function. The only magic is that it exhibits different limits in different regions.

- AravindG

That is why when we have absolute value function we give the answer in regions.

- anonymous

Mhmm i understand...

- anonymous

Im just thinking that I wouldn't have initially thought of approaching the problem like that (splitting it to 2 limits)

- 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

Properties like these are used by those who approach the problem rationally. It makes it more elegant.

- anonymous

Oh okay, that makes sense

- 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

That would eventually give me the same answer right?

- 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

You could do all these cases separately and evaluate it in your notebook to make it 100% clear.

- AravindG

That way you get more than one way to approach a problem.

- anonymous

Awesome! :D Thanks so much @AravindG !!! I bugged you a lot haha I'm sorry! :D

- AravindG

haha I like being bugged. It means the learner is a good student :)

- anonymous

Lol Thanks!! :D i appreciate it!

- AravindG

yw ;)

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