JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 3)

$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is equal to :
$${1 \over 3}$$
$${1 \over 4}$$
$${1 \over 6}$$
$${1 \over 12}$$

Explanation

$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}} \right)$$

$$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)$$

$$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)$$

$$ = {1 \over 6}$$

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