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