JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 14)
$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$ is equal to :
$${\pi ^2}$$
$$2{\pi ^2}$$
$$4{\pi ^2}$$
$$4\pi $$
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$$
$$ = {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$$
= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$
= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$$
$$ = {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$$
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