Question
The value of $$\mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt {1 + \cos \,x} }}$$ is :
A.
$$4\sqrt 2 {\left( {\log \,3} \right)^2}$$
B.
$$8\sqrt 2 {\left( {\log \,3} \right)^2}$$
C.
$$2\sqrt 2 {\left( {\log \,3} \right)^2}$$
D.
none of these
Answer :
$$8\sqrt 2 {\left( {\log \,3} \right)^2}$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt {1 + \cos \,x} }} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{9^x}{{.3}^x} - {9^x} - {3^x} + 1}}{{\sqrt 2 - \sqrt 2 \,\cos \frac{x}{2}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{9^x} - 1}}{x}} \right).\left( {\frac{{{3^x} - 1}}{x}} \right).\frac{1}{{\sqrt 2 }}.{x^2}.\frac{1}{{2\,{{\sin }^2}\frac{x}{4}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{9^x} - 1}}{x}} \right).\left( {\frac{{{3^x} - 1}}{x}} \right).\frac{1}{{\sqrt 2 }}\left( {\frac{{\frac{{{x^2}}}{{16}}}}{{{{\sin }^2}\frac{x}{4}}}} \right)8 \cr
& = \frac{8}{{\sqrt 2 }}\left( {\log \,9} \right)\left( {\log \,3} \right) \cr
& = 8\sqrt 2 {\left( {\log \,3} \right)^2} \cr} $$