JEE MAIN - Mathematics (2003 - No. 8)
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$$ is
$$\infty $$
$${1 \over 8}$$
0
$${1 \over 32}$$
Explanation
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {{\pi \over 4} - {x \over 2}} \right).\left( {1 - \sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$$
Let $$x = {\pi \over 2} + x;\,\,y \to 0$$
$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( {1 - \cos \,y} \right)} \over {{{\left( { - 2y} \right)}^3}}}$$
$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}$$
$$ = \mathop {\lim }\limits_{y \to 0} {1 \over {32}}{{\tan {y \over 2}} \over {\left( {{y \over 2}} \right)}}.{\left[ {{{\sin \,y/2} \over {y/2}}} \right]^2} = {1 \over {32}}$$
Let $$x = {\pi \over 2} + x;\,\,y \to 0$$
$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( {1 - \cos \,y} \right)} \over {{{\left( { - 2y} \right)}^3}}}$$
$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}$$
$$ = \mathop {\lim }\limits_{y \to 0} {1 \over {32}}{{\tan {y \over 2}} \over {\left( {{y \over 2}} \right)}}.{\left[ {{{\sin \,y/2} \over {y/2}}} \right]^2} = {1 \over {32}}$$
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