Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

^

Answer 5

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

Answer 6

And here is the reason it has provided: "Limit does not exist as LHL does not exist."

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

Answer 7

book is wrong

Answer 8

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

Answer 9

Oh okay. Thanks a lot bhaiya! :)

Answer 10

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

Answer 11

It is -1 to 1

[-1,1]

Answer 12

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

Answer 13

great, thanks again bhaiya

Answer 14

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

Answer 15

Yeah! I got it bhaiya...

Answer 16

that is a trick question actually, notice that x=1 is a boundary point

arcsinx is defined in [-1, 1]

Answer 17

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

Answer 18

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)

Answer 19

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?

Answer 20

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(+))

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

@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 ?

Answer 25

yea

Answer 26

so I can't understand your reasoning!

Answer 27

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

Answer 28

May be because LHL doesn't exist?

Answer 29

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