QJEEM19Mae244: JEE Main 2019 Mathematics - Class 12 JEEM19Mae244 Solution: equals:

Question

Verified Solution • 2025–26 Session moderate

\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}

equals:

Accepted Answer

\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

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