- mathslover

Evaluate \(\lim_{x\to 0} \sin^{-1} \left( \cfrac{\cos^{-1} x + \cos^{-1} (x^2) }{\pi} \right) \)

- jamiebookeater

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- ganeshie8

Hint :
\[\lim f(g(x)) = f(\lim g(x))\]
if \(f(x)\) is a continuous function

- mathslover

I don't know, I'm getting the answer as 0 .. but the book suggests that it doesn't exist :/

- Loser66

Is it not that when we replace x = 0, we have cos^(-1) (0) = pi/2, hence lim sin^(-1) (1) = pi/2?

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## More answers

- ganeshie8

^

- mathslover

Oh yeah, sorry for that. But, still, how does the limit not exist?

- mathslover

And here is the reason it has provided: "Limit does not exist as LHL does not exist."
\(\lim_{x \to 0^{-}} \left( \cfrac{\cos^{-1} x + \cos^{-1} (x^2)}{\pi} \right) = 1^{+} \)

- ganeshie8

book is wrong

- ganeshie8

http://www.wolframalpha.com/input/?i=lim%28x-%3E0%29+arcsin%28%28arccos%28x%29+%2B+arccos%28x%5E2%29%29%2Fpi%29

- mathslover

Oh okay. Thanks a lot bhaiya! :)

- ganeshie8

wait a second, maybe book is right... what is the domain of arccos(x) ?

- mathslover

It is -1 to 1
[-1,1]

- ganeshie8

Ahh right, we're good then. Limit exists and equals pi./2 :)

- mathslover

great, thanks again bhaiya

- ganeshie8

\[\large \lim\limits_{x\to 1} ~\arcsin(x) \stackrel{?}{ =}\frac{\pi}{2}\]

- mathslover

Yeah! I got it bhaiya...

- ganeshie8

that is a trick question actually, notice that x=1 is a boundary point
arcsinx is defined in [-1, 1]

- mathslover

Oh, but that will disturb the differentiability of the function, right? How's it going to affect the limit?

- ganeshie8

You're probably right, can you give me a convincing argument why the limit exists (recall that both side limits must exist and be same for the overall limit to exist)|dw:1434814029787:dw|

- mathslover

I see what you mean to point here. I'm just confused that we have arcsin(1) here, right? Then why will we take case of \(1^{+}\) here?

- mathslover

Even if we are considering RHL, then that doesn't mean we can simply put 1(+) at the place of 1 there...! right?
And the book shows that LHL is arcsin(1(+))

- ganeshie8

we simply don't worry about RHL here because the function itself is not defined for \(x\gt 1\)
a sequence can converge happily to a boundary point, no issue with it i guess
@zzr0ck3r @eliassaab

- anonymous

then why don't we do the same for
\[\sqrt{x} \]
?

- mathslover

Yeah, but here, x is tending to zero. Why will we take case of x >1 ? :/ Sorry if these questions sound stupid but..

- ganeshie8

@Catch.me are you suggesting \(\lim\limits_{x\to 0^{}}~\sqrt{x}\) doesn't exist if we think of \(\sqrt{x}\) as a real valued function ?

- anonymous

yea

- anonymous

so I can't understand your reasoning!

- ganeshie8

Okay :) may i knw why do you think \(\lim\limits_{x\to 0^{}}~\sqrt{x}\) doesn't exist ?

- mathslover

May be because LHL doesn't exist?

- misty1212

limit exists if
\[\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=L\]

- anonymous

because the function itself is not defined LHL

- ganeshie8

why do we care about what happens for x<0 where the function itself is not defined

- misty1212

if the function is not defined on one side, then a one sided limit exist, not limit

- anonymous

that is the definition of the limit x goes to 0
not + only or - only

- ganeshie8

that is valid only for the interior limit points

- misty1212

just part of the definition
there is no proviso that says "if the function does not exist to the right or left, then the one sided limit is the limit"

- ganeshie8

x=0 is a boundary point in our specific two cases. so we say the limit exists because we don't care what happens out side the domain

- anonymous

I think wolframalpha isn't concerned with only realvalues

- anonymous

so it gives a limit

- ganeshie8

i think it makes more sense if we consider sequences definition of limit, let me review it quick and get back

- ganeshie8

meantime here is a nice read
http://math.stackexchange.com/questions/637280/limit-of-square-root-of-x-as-x-approaches-0

- mathslover

^ That completely confused me. No one ever taught me about that :/ I was always taught that it doesn't exist

- anonymous

@ganeshie8 can you tell us what is equences definition of limit??

- ParthKohli

The limits of functions in terms of sequences? \[\lim_{x\to c} f(x) = L\]Just that if you consider all sequences \(\{ x_n\}\to c\) then the sequence \(\{f(x_n)\} \to L\)

- ParthKohli

And the limit of a sequence\[\lim x_n = c\]if there exists an \(M_0\) for all \(\epsilon > 0\) such that for all \(n > M_0\) we have \(|x_n- c|< \epsilon \)

- ganeshie8

thanks @ParthKohli , that looks neat!
I don't seem to remember much of real analysis :/

- mathslover

But still, I remain confused :(

- ParthKohli

Yeah, these definitions look completely nonsensical... until you draw this little picture. I guess you can use these to prove that\[\lim_{x\to 0}\sqrt x =0\]

- mathslover

Actually, I'm talking about the question at the top. Why does the limit not exist?

- ParthKohli

It doesn't?

- ganeshie8

just for the record-
for the main q, the limit exists and equals pi/2

- mathslover

Oh, thanks!

- ganeshie8

|dw:1434816607981:dw|

- ganeshie8

sequences can converge to a boundary point, we don't need to wry about what happens on the other side which is not part of the domain (other side simply doesn't exist as far as this particular problem is concerned)
|dw:1434816820615:dw|

- ParthKohli

Wow, that looks really nice.

- ganeshie8

that is as much real analysissy i can get, im actually looking for somebody else to explain it more better :)

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