mathslover
  • mathslover
Evaluate \(\lim_{x\to 0} \sin^{-1} \left( \cfrac{\cos^{-1} x + \cos^{-1} (x^2) }{\pi} \right) \)
Mathematics
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
Hint : \[\lim f(g(x)) = f(\lim g(x))\] if \(f(x)\) is a continuous function
mathslover
  • mathslover
I don't know, I'm getting the answer as 0 .. but the book suggests that it doesn't exist :/
Loser66
  • 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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ganeshie8
  • ganeshie8
^
mathslover
  • mathslover
Oh yeah, sorry for that. But, still, how does the limit not exist?
mathslover
  • 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
  • ganeshie8
book is wrong
ganeshie8
  • 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
  • mathslover
Oh okay. Thanks a lot bhaiya! :)
ganeshie8
  • ganeshie8
wait a second, maybe book is right... what is the domain of arccos(x) ?
mathslover
  • mathslover
It is -1 to 1 [-1,1]
ganeshie8
  • ganeshie8
Ahh right, we're good then. Limit exists and equals pi./2 :)
mathslover
  • mathslover
great, thanks again bhaiya
ganeshie8
  • ganeshie8
\[\large \lim\limits_{x\to 1} ~\arcsin(x) \stackrel{?}{ =}\frac{\pi}{2}\]
mathslover
  • mathslover
Yeah! I got it bhaiya...
ganeshie8
  • ganeshie8
that is a trick question actually, notice that x=1 is a boundary point arcsinx is defined in [-1, 1]
mathslover
  • mathslover
Oh, but that will disturb the differentiability of the function, right? How's it going to affect the limit?
ganeshie8
  • 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
  • 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
  • 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
  • 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
  • anonymous
then why don't we do the same for \[\sqrt{x} \] ?
mathslover
  • 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
  • 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
  • anonymous
yea
anonymous
  • anonymous
so I can't understand your reasoning!
ganeshie8
  • ganeshie8
Okay :) may i knw why do you think \(\lim\limits_{x\to 0^{}}~\sqrt{x}\) doesn't exist ?
mathslover
  • mathslover
May be because LHL doesn't exist?
misty1212
  • misty1212
limit exists if \[\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=L\]
anonymous
  • anonymous
because the function itself is not defined LHL
ganeshie8
  • ganeshie8
why do we care about what happens for x<0 where the function itself is not defined
misty1212
  • misty1212
if the function is not defined on one side, then a one sided limit exist, not limit
anonymous
  • anonymous
that is the definition of the limit x goes to 0 not + only or - only
ganeshie8
  • ganeshie8
that is valid only for the interior limit points
misty1212
  • 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
  • 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
  • anonymous
I think wolframalpha isn't concerned with only realvalues
anonymous
  • anonymous
so it gives a limit
ganeshie8
  • ganeshie8
i think it makes more sense if we consider sequences definition of limit, let me review it quick and get back
ganeshie8
  • ganeshie8
meantime here is a nice read http://math.stackexchange.com/questions/637280/limit-of-square-root-of-x-as-x-approaches-0
mathslover
  • mathslover
^ That completely confused me. No one ever taught me about that :/ I was always taught that it doesn't exist
anonymous
  • anonymous
@ganeshie8 can you tell us what is equences definition of limit??
ParthKohli
  • 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
  • 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
  • ganeshie8
thanks @ParthKohli , that looks neat! I don't seem to remember much of real analysis :/
mathslover
  • mathslover
But still, I remain confused :(
ParthKohli
  • 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
  • mathslover
Actually, I'm talking about the question at the top. Why does the limit not exist?
ParthKohli
  • ParthKohli
It doesn't?
ganeshie8
  • ganeshie8
just for the record- for the main q, the limit exists and equals pi/2
mathslover
  • mathslover
Oh, thanks!
ganeshie8
  • ganeshie8
|dw:1434816607981:dw|
ganeshie8
  • 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
  • ParthKohli
Wow, that looks really nice.
ganeshie8
  • 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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