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proof of the differential of sin x Watch

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      (Original post by Plato's Trousers)
      actually, yes, I was just re-reading it and wasn't happy with that part.
      Remember the law of limits: \lim \left[ f(h) g(h) \right] = \left[\lim f(h)\right] \left[\lim g(h) \right]
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      (Original post by Saichu)
      Remember the law of limits: \lim \left[ f(h) g(h) \right] = \left[\lim f(h)\right] \left[\lim g(h) \right]
      so do you mean that if we re-write -\dfrac{\sin^2 h}{h} as a product

      -\dfrac{\sin h}{h}\sin h

      and take the limits of each in turn, the product of those limits is the limit of the product?

      So,

      \displaystyle\lim_{h\rightarrow0  }-\dfrac{\sin h}{h}=1

      and


      \displaystyle\lim_{h\rightarrow0  }\sin h=0

      the the limit of the product is zero?
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        (Original post by Plato's Trousers)
        so do you mean that if we re-write as a product



        and take the limits of each in turn, the product of those limits is the limit of the product?

        So,



        and




        the the limit of the product is zero?
        that's correct
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        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0

        This is the problem line. The statement you need to make for your proof to work is:

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx

        You can then cancel the h on top by the h on the bottom. I know it's not a rigorous proof by any means, but it's what makes it work for C3/C4.

        It's like:
         \frac {0.0001 cos{x}}{0.0001} = cos{x}
         \frac {0.00000009999 cos{x}}{0.00000009999} = cos{x}

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0
        And
        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx
        Are equivalent on the limit..

        Try for yourself: sin(0.0001) x cos(x) = 0.000099999 cos)x)
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        (Original post by OS92)
        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0

        This is the problem line. The statement you need to make for your proof to work is:

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx
        Nope. His line was fine, it's your line which doesn't make sense.
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        (Original post by OS92)
        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0

        This is the problem line. The statement you need to make for your proof to work is:

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx

        You can then cancel the h on top by the h on the bottom. I know it's not a rigorous proof by any means, but it's what makes it work for C3/C4.

        It's like:
         \frac {0.0001 cos{x}}{0.0001} = cos{x}
         \frac {0.00000009999 cos{x}}{0.00000009999} = cos{x}

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0
        And
        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx
        Are equivalent on the limit..

        Try for yourself: sin(0.0001) x cos(x) = 0.000099999 cos)x)
        Sorry, but your whole post is nonsense. I don't think you really understand how limits work. In particular, this statement means nothing and is not true:

        \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx

        Once h tends to 0, it can't appear on the RHS.
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        (Original post by SimonM)
        Nope. His line was fine, it's your line which doesn't make sense.

        (Original post by Swayum)
        Sorry, but your whole post is nonsense. I don't think you really understand how limits work. [/latex]
        Thanks guys! I was starting to get confused again, having thought I understood it nicely
       
       
       
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