12. limits — MATHS STD 11 Science — Question

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\left( {x\tan 2x - 2x\tan \,x} \right)}}{{{{\left( {1 - \cos \,2x} \right)}^2}}} = \mathop {\lim }\limits_{x \to 0} K$ (say)

$ \Rightarrow K = \frac{{x\left[ {\frac{{2\tan \,x}}{{1 - {{\left( {\tan \,x} \right)}^2}}}} \right] - 2x\,\tan x}}{{{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$

$ = \frac{{2x\,\tan x - \left[ {2x\,\tan x - 2x\,{{\tan }^3}x} \right]}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ = \frac{{2x\,\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ \Rightarrow K = \frac{x}{{2\sin \,x \times \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\cos x}}$

$L = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}$

$ = \frac{1}{2}$