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anonymous

  • 5 years ago

I got a doubt in limits see below for my question

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  1. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow 0}\]\[sinax / tanbx\]

  2. anonymous
    • 5 years ago
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    its 0/0 form

  3. anonymous
    • 5 years ago
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    This is the actual question \[\lim_{x \rightarrow 0} sinax/tanbx\]

  4. anonymous
    • 5 years ago
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    differentiate top and bottom acos(ax) / b sec^2 (bx)

  5. anonymous
    • 5 years ago
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    sub x=0

  6. anonymous
    • 5 years ago
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    answer = (a/b)

  7. anonymous
    • 5 years ago
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    multiply and divide by bx/ax

  8. anonymous
    • 5 years ago
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    I don't know differntiation and u mean that i should multiply the numerator by ax/ax and denominator by bx/bx????

  9. anonymous
    • 5 years ago
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    I dont even know what him was on bout either lol

  10. anonymous
    • 5 years ago
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    if you havent done derivatives then your teacher shouldnt be giving you this question !

  11. anonymous
    • 5 years ago
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    this can be done without derivatives

  12. anonymous
    • 5 years ago
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    that's wat i am saying him1618 you are right

  13. anonymous
    • 5 years ago
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    how? , and its much easier with derivatives, 2lines of work

  14. anonymous
    • 5 years ago
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    its not pinching theorm

  15. anonymous
    • 5 years ago
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    yeah its easier..no doubt 110 times out of 100 id use derivatives, but the fact is that if hes doing limits fundamentally its imperative for him to know the fundamental steps..rather than shortcuts

  16. anonymous
    • 5 years ago
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    so i guess he wants a legitimate fundamental method which uses standard limit definitions

  17. anonymous
    • 5 years ago
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    exactly now can u pls explain

  18. anonymous
    • 5 years ago
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    first divide above and below by x

  19. anonymous
    • 5 years ago
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    then make a form like this \[[\sin( ax )/ax] /[\tan (bx)/bx] \] x a/b

  20. anonymous
    • 5 years ago
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    the book i am goin through says multiply the numerator by ax/ax and the denominator by bx/bx

  21. anonymous
    • 5 years ago
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    yeah thats right..dont look at the solns..which book is this?

  22. anonymous
    • 5 years ago
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    u understood wt i did?

  23. anonymous
    • 5 years ago
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    nope the book is my coaching book

  24. anonymous
    • 5 years ago
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    i did not understand wat wud happen if i divided by x????

  25. anonymous
    • 5 years ago
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    its the same thing

  26. anonymous
    • 5 years ago
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    ok divide sin ax by ax and tan bx by bx

  27. anonymous
    • 5 years ago
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    and then you will have to multiply by a/b also, to keep it the same..understand?

  28. anonymous
    • 5 years ago
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    so sinax by ax, with x tending to zero is 1, and so is tanbx by bx so the ans is 1 x a/b a/b

  29. anonymous
    • 5 years ago
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    samjha?

  30. anonymous
    • 5 years ago
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    yeh I remember that way , just the way the question looked put me off lol

  31. anonymous
    • 5 years ago
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    also, I just differentiate top and bottom etc, its so easy and there no brain required

  32. anonymous
    • 5 years ago
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    yeah everyone does that..me too.bt the kid wanted a method...i know no one in the world would not use derivatives for this..even the kid will start using em after a couple of months

  33. anonymous
    • 5 years ago
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    ok ab question yeh hai ki jo ax aur bx hum multiply aur divide karte hain woh idhar kya kar rahen hai???

  34. anonymous
    • 5 years ago
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    tujhe pata hai sinx / x is 1 when x tends to zero

  35. anonymous
    • 5 years ago
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    how do i k now dat is it a rule or smthing??

  36. anonymous
    • 5 years ago
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    its the first rule you learn in limits

  37. anonymous
    • 5 years ago
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    let me check

  38. anonymous
    • 5 years ago
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    sinx / x =1 and tanx / x is 1 when x tends to zero

  39. anonymous
    • 5 years ago
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    k yes got it!!!!!!

  40. anonymous
    • 5 years ago
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    so now if you have sin kx /x with x tending to zero and k any integer, whts the limit

  41. anonymous
    • 5 years ago
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    is it ax/bx??

  42. anonymous
    • 5 years ago
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    sin kx / x with x tending to zero??answer kya hai?

  43. anonymous
    • 5 years ago
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    1

  44. anonymous
    • 5 years ago
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    no its k

  45. anonymous
    • 5 years ago
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    multiply and divide by k

  46. anonymous
    • 5 years ago
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    yeah its k sorry yaar!!!!!!

  47. anonymous
    • 5 years ago
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    so in the same way u treat ur question as two different limits and then take their ratio...

  48. anonymous
    • 5 years ago
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    but if we do dis we get \[\sin ax/ax * ax = 1*ax = ax\]

  49. anonymous
    • 5 years ago
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    but then u have bx below..cancel x..u get a/b

  50. anonymous
    • 5 years ago
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    can i cancel (sin ax)/ax ka x???? to get a i don't think so!!!!

  51. anonymous
    • 5 years ago
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    vaise u cant..but when uve got b in the denominator u can

  52. anonymous
    • 5 years ago
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    how and y can i do dat??? any spl reason behind it???

  53. anonymous
    • 5 years ago
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    abe...............sinax / ax / tanbx / bx x a/b

  54. anonymous
    • 5 years ago
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    is it the same as sinax/tanbx

  55. anonymous
    • 5 years ago
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    yups

  56. anonymous
    • 5 years ago
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    to evaluate sinax/tanbx u need to form sinax/ ax/tanbx/bx....to make the expression same uve got to multiply by a/b

  57. anonymous
    • 5 years ago
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    luks like i got it the sin x / x = 1 wala way!!!!!!

  58. anonymous
    • 5 years ago
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    i got ax / bx where i can cancel x

  59. anonymous
    • 5 years ago
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    is this tareeka ok!!!!!!!!!

  60. anonymous
    • 5 years ago
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    sin ax / tan bx = sin(ax) x ax --------- ax ---------- tan (bx) x bx --------- bx

  61. anonymous
    • 5 years ago
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    theek hai?

  62. anonymous
    • 5 years ago
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    is the ax between the 2 xes for multiplying

  63. anonymous
    • 5 years ago
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    abe that x is multiply

  64. anonymous
    • 5 years ago
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    if so tab theek hai got it!!!!!!

  65. anonymous
    • 5 years ago
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    thanx got it!!!!!!!!

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