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Thread: Given function, evaluate limit?

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

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

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

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

    Differentiating w.r.t x

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

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

    Multiplying throughout by sinx

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

    $\displaystyle
    (1-\sin x)^2f(x)=\sin x
    $

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


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


    =e
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