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 May 13th, 2015, 08:01 AM #1 Newbie   Joined: May 2015 From: India Posts: 2 Thanks: 0 Limits [MATH] \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x} [\MATH] I took [MATH] \sin^2 x [\MATH] out of the brackets . Inside the brackets , I think I should use the formula [MATH] n(n-1)/2 [\MATH] . Am I doing right ? If yes, then what should I do next ? Thanks !
May 13th, 2015, 08:16 AM   #2
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To restate the problem:
Quote:
 Originally Posted by EllaRosewood $\displaystyle \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x}$ I took $\displaystyle \sin^2 x$ out of the brackets . Inside the brackets , I think I should use the formula $\displaystyle n(n-1)/2$ . Am I doing right ? If yes, then what should I do next ? Thanks !
$\tfrac12 n(n-1)$ is the formula for the sum of the first $n$ natural numbers (including zero). This is a harmonic series.

However, writing $s_n = \sum_{k=1}^n \tfrac1k$ we have
$$\left({s_n \over \sin^2 x}\right)^{\sin^2 x}$$
The numerator is heading to unity. So you need to investigate the limit of $x^{-x}$ as $x \to 0$. This is most easily achieved by writing $x^{-x} = \mathrm e^{-x \log x}$.

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