Question
$$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos \,2x} \right)\left( {3 + \cos \,x} \right)}}{{x\,\tan \,4x}}$$ is equal to-
A.
$$ - \frac{1}{4}$$
B.
$$\frac{1}{2}$$
C.
$$1$$
D.
$$2$$
Answer :
$$2$$
Solution :
Multiply and divide by $$x$$ in the given expression, we get
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos \,2x} \right)}}{{{x^2}}}.\frac{{\left( {3 + \cos \,x} \right)}}{1}.\frac{x}{{\tan \,4x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{2\,{{\sin }^2}\,x}}{{{x^2}}}.\frac{{3 + \cos \,x}}{1}.\frac{x}{{\tan \,4x}} \cr
& = 2\,\mathop {\lim }\limits_{x \to 0} \frac{{\,{{\sin }^2}\,x}}{{{x^2}}}.\mathop {\lim }\limits_{x \to 0} \,3 + \cos \,x.\mathop {\lim }\limits_{x \to 0} \,\frac{x}{{\tan \,4x}} \cr
& = 2.4.\frac{1}{4}\mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{\tan \,4x}} \cr
& = 2.4.\frac{1}{4} \cr
& = 2 \cr} $$