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Math Help - Given function, evaluate limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Given function, evaluate limit?

    If f(x) = \lim_{n\rightarrow \infty} \left\{\sin x + 2\sin^2 x + 3\sin^3 x + \mbox{...} + n\sin^n x\right\}, then evaluate:

    \lim_{x\rightarrow \frac{\pi}{2}} \left\{(1 - \sin x)^2 f(x)\right\}^{\frac{1}{(sin x - 1)}}.
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  2. #2
    Senior Member
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    Telescoping

    Well, first notice that (1-\sin x)^2f(x) is a telescoping series converging on simply \sin x. So to evaluate L=\lim_{x\rightarrow \frac\pi2} (\sin x)^{\frac1{\sin x-1}} , we first substitute y=\sin x , making L=\lim_{y\rightarrow0}y^{\frac1{y-1}} and then z=\frac1{y-1}, so L=\lim_{z\rightarrow\infty}(1+\frac1z)^z=e
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  3. #3
    Senior Member pankaj's Avatar
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    1+\sin x+\sin^2 x+\sin^3 x+.........=\frac{1}{1-\sin x}

    Differentiating w.r.t x

    \cos x+2\sin x\cos x+3\sin^2 x\cos x+4\sin^3 x\cos x+.....=\frac{\cos x}{(1-\sin x)^2}

    1+2\sin x+3\sin^2 x+4\sin^3 x+.....=\frac{1}{(1-\sin x)^2}

    Multiplying throughout by sinx

     <br />
\sin x+2\sin^2 x+3\sin^3 x+4\sin^4 x+.....=\frac{\sin x}{(1-\sin x)^2}<br />

     <br />
(1-\sin x)^2f(x)=\sin x<br />

    \lim_{x\to \frac{\pi}{2}}((1-\sin x)^2f(x))^{\frac{1}{\sin x-1}}<br />


    =e^{\lim_{x\to \frac{\pi}{2}}((1-\sin x)^2f(x)-1){\frac{1}{\sin x-1}}}


    =e
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