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anonymous

  • 5 years ago

use l'hopital's rule to solve the limit as it approaches 0 from the right of 3/sinx - 1/x ??

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  1. anonymous
    • 5 years ago
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    \[ \lim_{x \rightarrow 0+} 3/\sin x - 1/x\]

  2. anonymous
    • 5 years ago
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    limit is 0 l'hopitals rule you'll get lim of 0/cos(0) - 0/1 so 0

  3. anonymous
    • 5 years ago
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    so 0 is also an indeternminate form?

  4. anonymous
    • 5 years ago
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    thanks by the wayyyy!!! :)

  5. anonymous
    • 5 years ago
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    no, if the limit is 0, it exists it goes to 0

  6. anonymous
    • 5 years ago
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    if it comes out to be 0 / 0 then it's indeternminate

  7. anonymous
    • 5 years ago
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    First show that limit is indeterminate, then employ l'hopital's rule lim x->0 (3x-sinx)/(xsinx)

  8. anonymous
    • 5 years ago
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    i'm not sure if cyter has it right... i'm pretty sure you deriv instead of integral for l'hopital's rule

  9. anonymous
    • 5 years ago
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    this function ends up equaling 0.. thats where i am confused bcuz why do i need to use l'hopitals rule/.

  10. anonymous
    • 5 years ago
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    lim of 0/cos(0) - 0/1 ends up zero so the limit is zero but your original equation (the question) is undefined (or indeter.) so you have to apply l'hoptials' rule

  11. anonymous
    • 5 years ago
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    It's not 0. You can't use (and don't need) l'Hopital here.

  12. anonymous
    • 5 years ago
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    polpak, can you explain your reasoning?

  13. anonymous
    • 5 years ago
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    Err wait. It's not 0, but you do need l'hopital

  14. anonymous
    • 5 years ago
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    how is my original equation (the question) undefined? 3 over sin(0) - 1 over x = 3 over 0 - 1 over 0 = 0 - 0 = 0 ??

  15. myininaya
    • 5 years ago
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    polpak when you get a chance will you come back to that kid who is trying to teach himself

  16. anonymous
    • 5 years ago
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    3/0 is not 0 it's undefined. which is why i would do l'hopitals rule but polpak says i'm wrong :x so we'll see what he says first

  17. anonymous
    • 5 years ago
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    You don't have 0 - 0, you have \(3/0 - 1/0 = \infty - \infty\)

  18. anonymous
    • 5 years ago
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    And you do need lhopital, but you need to first put over a common denominator

  19. anonymous
    • 5 years ago
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    ohhhh *faceplam* i forgot limits >< it's "as you approach" not zero thanks polpak yup itenn, polpak's right

  20. anonymous
    • 5 years ago
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    \[3/sinx - 1/x = \frac{3x - sinx}{xsinx} \] Which gives a 0 over 0 and you take the derivative of top and bottom.

  21. anonymous
    • 5 years ago
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    Give that derivative a try and take the limit as the derivative aproaches 0

  22. anonymous
    • 5 years ago
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    I'll brb

  23. anonymous
    • 5 years ago
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    thankk you!!!!

  24. anonymous
    • 5 years ago
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    so far i have 3 - cos x all over 2 cos x - x sinx

  25. anonymous
    • 5 years ago
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    I agree with the numerator, but not the denominator.

  26. anonymous
    • 5 years ago
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    Using products rule \[\frac{d}{dx}x(sin\ x) = x(cos\ x) + (sin\ x)\]

  27. anonymous
    • 5 years ago
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    Which gives \[\frac{3-cosx}{x(cos\ x) + (sin\ x)}\] Which we can then plug in 0 for x

  28. anonymous
    • 5 years ago
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    2/0 ?

  29. anonymous
    • 5 years ago
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    Yep, \(2/0^+\)

  30. anonymous
    • 5 years ago
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    So \(+\infty\)

  31. anonymous
    • 5 years ago
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    why was the first equation indeterminate again.. im sorry im not getting this:(

  32. anonymous
    • 5 years ago
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    it was similat to the 2/0+ ... it was 3/0 - 1/0.

  33. anonymous
    • 5 years ago
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    Because you had 3/0 - 1/0. Which is \(\infty - \infty\)

  34. anonymous
    • 5 years ago
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    And that can be a lot of different things. It could be 0, it could be a constant, or it could be +/- infinity.

  35. anonymous
    • 5 years ago
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    So we have to put them over a common denominator to get 0/0 or \(\infty/\infty\) and then use l'Hopital's

  36. anonymous
    • 5 years ago
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    to see which infinity is growing faster

  37. anonymous
    • 5 years ago
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    ohkay. i did another on my own -- limit x approaches 1 [ x^3 - 1 all over x-1 ]

  38. anonymous
    • 5 years ago
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    i got 0/0 and then did derivative of top and bottom 3x^2 over 1

  39. anonymous
    • 5 years ago
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    and then i got 3. is that right?

  40. anonymous
    • 5 years ago
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    Yep, that's right.

  41. anonymous
    • 5 years ago
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    my last problem for l'hopitals rule is limit x approaches 0+ x^x ??

  42. anonymous
    • 5 years ago
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    what is the indeterminate form... if 0^0 = 0

  43. anonymous
    • 5 years ago
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    It's not 0. If you have \(\lim_{x \rightarrow c} f(x) = 0^0\) then you have to take the \[\lim_{e^x \rightarrow c} ln\ f(x)\]

  44. anonymous
    • 5 years ago
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    So in this case you have \[\lim_{e^x \rightarrow 0} x(ln\ x)\] Which is of the form \(0\cdot\infty\)

  45. anonymous
    • 5 years ago
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    0^0 is actually an indeterminate form so u can use l'hopital's rule

  46. anonymous
    • 5 years ago
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    and 0^0 is not equal to 0

  47. anonymous
    • 5 years ago
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    0 * inf is an indeterminate form?

  48. anonymous
    • 5 years ago
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    no

  49. anonymous
    • 5 years ago
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    polpk, I have a quick question on mine you were helping me with.. when you're done here:)

  50. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

  51. anonymous
    • 5 years ago
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    inf/inf, 0/0, -inf/inf, inf/-inf or 0^0, any of those are indeterminate forms

  52. anonymous
    • 5 years ago
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    \(0 * \infty\) is indeterminate.

  53. anonymous
    • 5 years ago
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    sometimes wikipedia is unreliable

  54. anonymous
    • 5 years ago
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    the next step then would be to take the derivativve?

  55. anonymous
    • 5 years ago
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    but after futhre digging I'm convinced

  56. anonymous
    • 5 years ago
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    Was gonna say, wolfram alpha also thinks it's indeterminate: http://mathworld.wolfram.com/Indeterminate.html

  57. anonymous
    • 5 years ago
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    Which next step? I may have a different process for this than cyter.

  58. anonymous
    • 5 years ago
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    im following you polpak..

  59. anonymous
    • 5 years ago
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    I'm not sure how he was gonna use l'hopital on 0^0 since there's no fraction there, but I'm a bit rusty with this

  60. anonymous
    • 5 years ago
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    thats where i confused.. l'hopitals rule has lim f(x)/g(x) and then f'(x)/g'(x)

  61. anonymous
    • 5 years ago
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    i dont see a fraction here.

  62. anonymous
    • 5 years ago
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    Since \[\lim_{e^x \rightarrow 0} x(ln\ x)\] is not nice, we have to swap our variable around a bit and say u = 1/x if x goes to 0, u goes to infinity. And that means that x = 1/u, so we have now \[\lim_{e^{u} \rightarrow \infty} \frac{ln\ 1/u}{u}\] Which gives us a nice fraction with \(\infty/\infty\) that we can take the derivative of.

  63. anonymous
    • 5 years ago
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    this one is very complicated.

  64. anonymous
    • 5 years ago
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    Yeah, a bit.

  65. anonymous
    • 5 years ago
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    i am not good with derivatices so the ln is trouble for me.. this is hard:(

  66. anonymous
    • 5 years ago
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    So \[\frac{d}{du} ln\ 1/u = u*(\frac{-1}{u^2}) = \frac{-1}{u}\]

  67. anonymous
    • 5 years ago
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    ohhh okay.

  68. anonymous
    • 5 years ago
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    so now we have Erg, my notation is wrong, but it's tiny if I fix it.

  69. anonymous
    • 5 years ago
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    what would change?

  70. anonymous
    • 5 years ago
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    Anyway, we have It's supposed to be e raised to the power of the limit of that whole thing, not e^u approaches infinity.

  71. anonymous
    • 5 years ago
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    So we have \[e^{\lim_{u \rightarrow \infty} \frac{-1}{t}} = e^0 = 1\]

  72. anonymous
    • 5 years ago
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    See, tiny.

  73. anonymous
    • 5 years ago
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    mm.. at which point?

  74. anonymous
    • 5 years ago
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    ohh it was never supposed to be e to the u

  75. anonymous
    • 5 years ago
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    Yeah, it was supposed to be e to the limit. That's it. The limit as \(x \rightarrow 0 \text{ of } x^x = 1\)

  76. anonymous
    • 5 years ago
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    so no l'hopital is needed??

  77. anonymous
    • 5 years ago
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    yes it is. I used it.

  78. anonymous
    • 5 years ago
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    This is too difficult with such a tiny font. Lemme try something else

  79. anonymous
    • 5 years ago
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    ohkay.

  80. anonymous
    • 5 years ago
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    i am confused:(

  81. anonymous
    • 5 years ago
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    http://www.dabbleboard.com/draw?b=Guest665733&i=0&c=68e3fba29b21f18d6667a1c196154c7ed4b665ee

  82. anonymous
    • 5 years ago
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    im there..

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