Question

$L{t}_{x\to 0}\frac{\exp \left[\right({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \infty \left){\log }_{e}2\right]-1}{x^2}$ equals

• A. ${\log }_{e}2$
• B. $2{\tan }^{2}x$
• C. $0$
• D. $1$

Answer 1

The correct option is A ${\log }_{e}2$

First we simplify the expression in powers of $\sin x$
${\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+..........+\infty$
This is an infinite GP with first term ${sin}^{2}x$ and common ratio ${\sin }^{2}x$
Hence it's sum will be,
$\frac{{\sin }^{2}x}{1-{\sin }^{2}x}$
$=\frac{{\sin }^{2}x}{{\cos }^{2}x}$
$={\tan }^{2}x$
The question simplifies to,
lim $x\to 0\frac{exp\left({\tan }^{2}x\log 2\right)-1}{x^2}$
= lim $x\to 0\frac{2^{{\tan }^{2}x}-1}{x^2}$
Multiplying and dividing by ${\tan }^{2}x$
= lim $x\to 0\frac{2^{{\tan }^{2}x}-1}{{\tan }^{2}x}\times \frac{{\tan }^{2}x}{x^2}$
Using the respective identity,
$=1\times 1$
$=1$