A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mathslover

  • one year ago

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

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  2. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    ^

  5. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    book is wrong

  8. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh okay. Thanks a lot bhaiya! :)

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  10. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  12. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    great, thanks again bhaiya

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  14. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah! I got it bhaiya...

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  16. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    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|

  18. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  19. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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(+))

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    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

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  22. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    @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 ?

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so I can't understand your reasoning!

  26. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  27. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    May be because LHL doesn't exist?

  28. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because the function itself is not defined LHL

  30. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  31. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    that is valid only for the interior limit points

  34. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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"

  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    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

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think wolframalpha isn't concerned with only realvalues

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so it gives a limit

  38. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  39. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  40. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  42. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \(\{f(x_n)\} \to L\)

  43. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \(|x_n- c|< \epsilon \)

  44. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  45. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But still, I remain confused :(

  46. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  47. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  48. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It doesn't?

  49. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  50. mathslover
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, thanks!

  51. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    |dw:1434816607981:dw|

  52. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    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|

  53. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wow, that looks really nice.

  54. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

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

  55. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.