Question

Solve:

$\underset{x\to 0}{lim}{(\frac{1}{2{\sin }^{2}x}+\frac{1}{3{\sin }^{2}x}+\frac{1}{4{\sin }^{2}x}+......\frac{1}{20{\sin }^{2}x})}^{{\sin }^{2}x}$

Answer 1

$\underset{x\to 0}{lim}{(\frac{1}{2{\sin }^{2}x}+\frac{1}{3{\sin }^{2}x}+\frac{1}{4{\sin }^{2}x}+......\frac{1}{20{\sin }^{2}x})}^{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}{\left(\frac{1}{{\sin }^{2}x}[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......\frac{1}{20}]\right)}^{{\sin }^{2}x}$

$=2.598\underset{x\to 0}{lim}{\left(\frac{1}{{\sin }^{2}x}\right)}^{{\sin }^{2}x}$

$=2.598\underset{x\to 0}{lt}{e}^{ln{\left(\frac{1}{{\sin }^{2}x}\right)}^{{\sin }^{2}x}}$

$=2.598e^{\underset{x\to 0}{lt}{\sin }^{2}xln\left(\frac{1}{{\sin }^{2}x}\right)}$

$=2.598e^{\underset{x\to 0}{lt}\frac{\left({cosec}^{2}x\right)}{{cosec}^{2}x}}$

$=2.598e^{\underset{x\to 0}{lt}\frac{1}{{cosec}^{2}x}.\frac{\frac{d}{dt}{cosec}^{2}x}{\frac{d}{dt}{cosec}^{2}x}}$

$=2.598e^{\underset{x\to 0}{lt}{\sin }^{2}x}$

$\simeq 2.598$