Question
The integer $$n$$ for which $$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\cos \,x - 1} \right)\left( {\cos \,x - {e^x}} \right)}}{{{x^n}}}$$ is a finite non-zero number is-
A.
$$1$$
B.
$$2$$
C.
$$3$$
D.
$$4$$
Answer :
$$3$$
Solution :
Given that,
$$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\cos \,x - 1} \right)\left( {\cos \,x - {e^x}} \right)}}{{{x^n}}} = $$ finite non zero number
$$\eqalign{
& \mathop { = \lim }\limits_{x \to 0} \frac{{\left( {1 - \cos \,x} \right)\left( {1 + \cos \,x} \right)\left( {{e^x} - \cos \,x} \right)}}{{{x^n}\left( {1 + \cos \,x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sin }^2}\,x}}{{{x^2}}}} \right).\left( {\frac{{{e^x} - \cos \,x}}{{{x^{n - 2}}}}} \right).\left( {\frac{1}{{1 + \cos \,x}}} \right) \cr
& = \mathop {\lim }\limits_{x \to 0} {1^2}.\frac{{{e^x} - \cos \,x}}{{{x^{n - 2}}}}.\frac{1}{2} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \sin \,x}}{{\left( {x - 2} \right){x^{n - 3}}}}\,\,\,\left[ {{\text{using L'Hospital rule}}} \right] \cr} $$
For this limit to be finite, $$n - 3 = 0 \Rightarrow n = 3$$