1. Standard membergenius
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    20 Aug '03 19:32
    differentiate cos2x from first principles. so then, why does
    lim
    h->0 ((sin2h)/h)=2 ?...

    i was typing it all out then lost it all and couldn't be bothered typing it up again...😛
  2. DonationAcolyte
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    20 Aug '03 20:00
    Er... what's your definition of sin? If it's the power series, then it's because all the terms of sin2h/h tend to zero except 2h/h, which is 2.

    I get the feeling I'm missing something here 😕
  3. Standard membergenius
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    20 Aug '03 20:101 edit
    this isn't so much a *problem* as a well, problem 😛 it came up in maths the other day, and my teacher anounced that e had no idea why this was, and neither did anyone else in the maths department...and i have no idea what you mean about power series and sin 😕
  4. Standard memberroyalchicken
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    20 Aug '03 21:00
    Hey genius. d(sin 2h)/dh = 2cos2h. Since sin2h--->0 as h--->0, and h--->0 as h--->0, just apply l'Hopital's rule and you get:

    lim sin2h/h = 2 lim cos2h = 2.

  5. Standard membergenius
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    20 Aug '03 21:08
    Originally posted by royalchicken
    Hey genius. d(sin 2h)/dh = 2cos2h. Since sin2h--->0 as h--->0, and h--->0 as h--->0, just apply l'Hopital's rule and you get:

    lim sin2h/h = 2 lim cos2h = 2.

    l'Hopital's rule?...😛
  6. Standard memberroyalchicken
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    20 Aug '03 21:14
    'Tis fairly standard, and it explains this. If the limit would be an indeterminate form of the right type (inf/inf, 0/0, 0*inf), then:

    lim f(x)/g(x) = lim f'(x)/g'(x)

    with as many repetitions as needed.
  7. Standard memberroyalchicken
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    21 Aug '03 00:29
    Originally posted by Acolyte
    Er... what's your definition of sin?
    Acolyte, I do believe this belongs in the thread on God😉
  8. Standard memberFiathahel
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    21 Aug '03 08:21
    You can also use Taylor.
    f(x) = f(0) + x * f'(0) + 1/2 x^2 f''(q) for some q in [0,x]

    then sin x = sin 0 + x cos 0 + 1/2 x^2 sin q = x + 1/2 x^2 sin q

    then (sin x)/x = 1 + 1/2 x sin q --> 1 if x-->0

    sin(2h)/h = (2 sin h cos h)/h = 2 (sin h)/h cos h = 2 * 1 * 1 = 2
  9. Standard memberroyalchicken
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    21 Aug '03 16:29
    Indeed you can. Good show. I think this was mentioned before, and indeed I prefer to think of the circular functions in terms of their Maclaurin series, because for complex arguments the standard definition of say, sin z becomes rather meaningless.
  10. DonationAcolyte
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    23 Aug '03 09:33
    Originally posted by royalchicken
    Acolyte, I do believe this belongs in the thread on God😉
    3 points to RC... sorry, wrong thread.
  11. Standard memberroyalchicken
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    23 Aug '03 20:40
    Colin, you've used the wrong thread. Take two cards.

    ~Chairman Chicken 😛
  12. Donation!~TONY~!
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    24 Aug '03 16:311 edit
    You know guys, sometimes I think you just post bogus stuff to make everyone else that comes in here feel so stupid! 🙄 Hahahaha, all these rules I have never heard of and all these nuts equations! My brain starts wimpering on the spot when I see that stuff! Hahahaha, Just Kidding guys. I'm actually jealous that I can't help out with this stuff! Carry on. 😀
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