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Math Help - Limit of trigonometric function

  1. #1
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    Exclamation Limit of trigonometric function

    Can I get some help in finding the limit of the following trigonometric function?

    Lt sin(a+3h)-3sin(a+h)+3sin(a+h)-sin a
    h-->0 h^3



    the h in the denominator is cube.


    the denominator is sin(a+3h)-3sin(a+2h)+3sin(a+h)-sin a
    Last edited by tariq_h_tauheed; September 13th 2008 at 02:42 PM. Reason: asked by someone.
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  2. #2
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    Quote Originally Posted by tariq_h_tauheed View Post
    Can I get some help in finding the limit of the following trigonometric function?

    Lt sin(a+3h)-3sin(a+h)+3sin(a+h)-sin a
    h-->0 h^3
    Is your question like this:

    \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - 3\sin \left( {a + h} \right) + 3\sin \left( {a + h} \right) - \sin a}}<br />
{{h^3 }}<br />
\

    or is your question like this:

    \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right)  - \sin a}}<br />
{{h^3 }}<br />
\
    Last edited by Shyam; September 13th 2008 at 01:43 PM.
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  3. #3
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    Hello, tariq_h_tauheed!

    Where did this problem come from?
    It looks like we walked in on the middle of someone's work . . .



    \lim_{h\to0}\frac{\sin(a+3h) - \overbrace{3\sin(a+h) + 3\sin(a+h)}^{\text{Don't these cancel?}} - \sin(h)}{h^3}

    Could it be: . \frac{\sin(a+3h) - \sin({\color{red}a})}{h^3}

    If so, we have a chance of solving it . . . otherwise, forget it!

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