JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 12)
$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$ equals:
$$ \sqrt 2$$
$$2 \sqrt 2$$
4
$$4 \sqrt 2$$
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {1 - \cos x}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $$\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $${1 \over 2} \times 4 \times 2\sqrt 2 $$
= $$4\sqrt 2 $$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {1 - \cos x}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $$\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$$
= $${1 \over 2} \times 4 \times 2\sqrt 2 $$
= $$4\sqrt 2 $$
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