Question
$$\mathop {\lim }\limits_{x\, \to \,0} \frac{{x\,\tan \,2x - 2x\,\tan \,x}}{{{{\left( {1 - \cos \,2x} \right)}^2}}}$$ is-
A.
$$2$$
B.
$$ - 2$$
C.
$$ \frac{1}{2}$$
D.
$$ - \frac{1}{2}$$
Answer :
$$ \frac{1}{2}$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x\, \to \,0} \frac{{x\,\tan \,2x - 2x\,\tan \,x}}{{{{\left( {1 - \cos \,2x} \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x\, \to \,0} \frac{{x\left\{ {2x + \frac{{8{x^3}}}{3} + \frac{{64{x^5}}}{{15}} + ...} \right\} - 2x\left\{ {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ...} \right\}}}{{4{{\sin }^4}x}} \cr
& = \mathop {\lim }\limits_{x\, \to \,0} \frac{{{x^4}\left\{ {\frac{8}{3} - \frac{2}{3} + {\text{ terms containing higher positive powers of }}x} \right\}}}{{4{{\sin }^4}x}} \cr
& = \frac{1}{4}.2 \cr
& = \frac{1}{2} \cr} $$