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jatinbansalhot

  • 2 years ago

find out the limit of the following ( sin x ) / x where x -> 0

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  1. ankit709
    • 2 years ago
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    =1

  2. jatinbansalhot
    • 2 years ago
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    plz explain too . i found out it to be 0. u r right it is 1 how

  3. furan143
    • 2 years ago
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    answer is 1,actually it's proof is a bit theoretical .if you want any proof please refer some standard text

  4. furan143
    • 2 years ago
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    it's proof is well known and is known as sandwich theorem

  5. jatinbansalhot
    • 2 years ago
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    as x reduces, the sin of x also reduces, then why 1

  6. ankit709
    • 2 years ago
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    if u put the value of limit then u will get 0/0 form

  7. jatinbansalhot
    • 2 years ago
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    i will get 0

  8. furan143
    • 2 years ago
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    offcourse buddy

  9. ankit709
    • 2 years ago
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    u can apply dy/dx

  10. furan143
    • 2 years ago
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    actually 0/0 is a indeterminate value

  11. jatinbansalhot
    • 2 years ago
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    derivative ? but i don't know the derivative i'm beginner but soon be older :D :D hahahaa

  12. ankit709
    • 2 years ago
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    this is L hospital rule

  13. jatinbansalhot
    • 2 years ago
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    furan.. yah that would be indeterminate state

  14. jatinbansalhot
    • 2 years ago
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    ohhh.. hospital ....

  15. furan143
    • 2 years ago
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    I think the proof starts from some sort of triangle & using some inequality conditions

  16. furan143
    • 2 years ago
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    buddy L'hospital rule is not the actual prrof

  17. furan143
    • 2 years ago
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    srry proof

  18. ankit709
    • 2 years ago
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    I KNOW but u also know that proof of this equation is 2 long

  19. jatinbansalhot
    • 2 years ago
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    i think for the time being, it is enough that the limit of that is = 1 .... as i will advance in this calculus, i will come across the proof soon and become familiar with that ...

  20. jatinbansalhot
    • 2 years ago
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    really too long ? i thought it would be very simple

  21. furan143
    • 2 years ago
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    okay then you better use L'hospital's rule then

  22. jatinbansalhot
    • 2 years ago
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    yah.... i will ... :) thanks all frnds in here .. :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

  23. furan143
    • 2 years ago
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    actually sandwich theorem is like the foundation of calculus

  24. furan143
    • 2 years ago
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    proof is not that long but concept is difficult

  25. jatinbansalhot
    • 2 years ago
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    sandwich.... i need to work up at great deal :) :) i will .. bcoz i want to be so perfect in calculus :)

  26. furan143
    • 2 years ago
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    I remembers the last step

  27. jatinbansalhot
    • 2 years ago
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    it's alright no problem. thanks you all alot...... really helpful for me.... :)

  28. CeW
    • 2 years ago
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    sin(x) for small x is x. x/x is 1 for any x.

  29. furan143
    • 2 years ago
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    cosx<(sinx/x)<1 here (sinx/x) is sandwiched between cosx & 1 (here not that cosx is also 1 when x tends to zero so sinx/x has no other option other than to tend to zero) that's why it has got the name sandwich or squeeze theorem

  30. CeW
    • 2 years ago
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    sin(x) / x does not approach 0 in the limit of x -> 0 !!

  31. furan143
    • 2 years ago
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    you are right cew

  32. jatinbansalhot
    • 2 years ago
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    santosh r u there?

  33. experimentX
    • 2 years ago
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    yeah

  34. jatinbansalhot
    • 2 years ago
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    hi experimentX, how are you brooo...

  35. experimentX
    • 2 years ago
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    doing fine ... i might be offline pretty soon .. |dw:1334192708048:dw|

  36. jatinbansalhot
    • 2 years ago
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    before being, could you expect to have some time with me or not ?

  37. jatinbansalhot
    • 2 years ago
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    open these three :)

  38. experimentX
    • 2 years ago
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    you see arc there = it's value = \(r\theta\), when \( \theta->0\), the ARC and H seem to be one line. or h/rQ -> tends to one as Q->0 h/r = sinQ => sinQ/Q->1 as Q->0

  39. jatinbansalhot
    • 2 years ago
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    i want to know why epsilon is smaller than a - x0 and x1 - a

  40. jatinbansalhot
    • 2 years ago
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    yah.... both approach toward 0 i saw that

  41. experimentX
    • 2 years ago
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    It's not both approach to zero ... it's how their ratio approach as Q approaches to zero -> that is limit.

  42. jatinbansalhot
    • 2 years ago
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    why epsilon is smaller than a - x0 and x1 - a ............................... :)

  43. experimentX
    • 2 years ago
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    how is epsilon defined??

  44. jatinbansalhot
    • 2 years ago
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    i know how epsilon is defined but why epsilon is smaller than a - x0 and x1 - a ............................... :)

  45. experimentX
    • 2 years ago
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    i don't know how epsilon is defined ... is it |x-a| < epsilon or |f(x) - L | < epsilon

  46. jatinbansalhot
    • 2 years ago
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    omggggggggggggg... how i can get answer to my question... here no one knows about limit definition properly..... ... leave it i try to google it... :D

  47. jatinbansalhot
    • 2 years ago
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    thanks experimentX..... :) :)

  48. jatinbansalhot
    • 2 years ago
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    ExperimentX... I go and study mathematics more and more if again any ques i would come back :) take care brooo :)

  49. experimentX
    • 2 years ago
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    sure ...

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