Question 309 of 401MediumMCQ
\[\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}\] equals :
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Official solution
Let, L = \[\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}}\] = \[\mathop {\lim }\limits_{x \to 0} K\] (say)
\[ \Rightarrow \] K = \[{{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}\]
= \[{{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}\]
= \[{{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}\]
= \[{{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}\]
= \[{{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}\]
\[ \Rightarrow \]K = \[{x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}\]
\[\therefore\,\,\,\] L = \[\mathop {\lim }\limits_{x \to 0} \] \[{x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}\]
= \[\mathop {\lim }\limits_{x \to 0} \] \[{x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}\]
= \[{1 \over 2}\]
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