Question
$$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \,x}}{{x\left( {{2^x} - 1} \right)}}$$ is equal to :
A.
$$\frac{1}{2}\,{\log _2}e$$
B.
$$\frac{1}{2}\,{\log _e}2$$
C.
1
D.
none of these
Answer :
$$\frac{1}{2}\,{\log _2}e$$
Solution :
$$\eqalign{
& {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \,x}}{{{x^2}}}.\frac{x}{{{2^x} - 1}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{2\,{{\sin }^2}\frac{x}{2}}}{{4.{{\left( {\frac{x}{2}} \right)}^2}}}.\frac{x}{{{e^{x\,{{\log }_e}2}} - 1}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{2}.{\left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2}.\frac{x}{{\left\{ {1 + x\,{{\log }_e}2 + \frac{{{{\left( {x\,{{\log }_e}2} \right)}^2}}}{{n!}} + .....} \right\} - 1}} \cr
& = \frac{1}{2}.\frac{1}{{{{\log }_e}2}} \cr
& = \frac{1}{2}{\log _2}e\, \cr} $$