QJEEM22Ma5258: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5258 Solution

Let $x + 2\cos x = a$

$x + 2 = b$

as $x \to 0$ , $a \to 2$ and $b \to 2$

$\mathop {\lim }\limits_{x \to 0} {\left( {{{{a^3} + 2{a^2} + 3\sin a} \over {{b^3} + 2{b^2} + 3\sin b}}} \right)^{{{100} \over x}}}$

$= {e^{\mathop {\lim }\limits_{x \to 0} \,.\,{{100} \over x}\,.\,{{({a^3} - {b^3}) + 2({a^2} - {b^2}) + 3(\sin a - \sin b)} \over {{b^3} + 2{b^2} + 3\sin b}}}}$

$\because$ $\mathop {\lim }\limits_{x \to 0} {{a - b} \over x} = \mathop {\lim }\limits_{x \to 0} {{2(\cos x - 1)} \over x} = 0$

$= {e^0}$

$= 1$