Limits

May 2015
2
0
India
\(\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} [\MATH]

I took \(\displaystyle \sin^2 x [\MATH] out of the brackets . Inside the brackets , I think I should use the formula \(\displaystyle n(n-1)/2 [\MATH] . Am I doing right ? If yes, then what should I do next ? Thanks !\)\)\)
 

v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
To restate the problem:
\(\displaystyle \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x} \)

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 $x^{-x} = \mathrm e^{-x \log x}$.