Question

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

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

3/2? I am getting 1/4

Answer 6

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

Answer 7

Yeah i mean here...

\[\frac{ 3 }{ 2 } \lim _{x \rightarrow1}\frac{ x-1 }{ \left| x-1 \right| }\]

Answer 8

where does the 3/2 come from?

Answer 9

Is this the same question?

Answer 10

yeah, check the solution in the attachment

Answer 11

Ohh my bad wrong attachment haha wait a sec

Answer 12

I dont see 3/2

Answer 13

:|

Answer 14

in this one XD sorry lol

Answer 15

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

Answer 16

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

Answer 17

They split the limit as

lim xto1 (x+2)/(x+1) times lim xto1 x-1/|x-1|

Answer 18

lim(g(x)f(x))=lim(g(x)) times lim(f(x))

Thats what they used

Answer 19

Get it?

Answer 20

Ohhh yeah now i get it

Answer 21

well done :)

Answer 22

but why?
since there's an absolute value?

Answer 23

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).

Answer 24

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

Answer 25

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

Answer 26

Nope.

Answer 27

and just get 3/2

Answer 28

|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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Mhmm i understand...

Answer 34

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

Answer 35

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.

Answer 36

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

Answer 37

Oh okay, that makes sense

Answer 38

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?

Answer 39

That would eventually give me the same answer right?

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

Lol Thanks!! :D i appreciate it!

Answer 46

yw ;)