A) \[\sqrt{5}\,{{(\log 3)}^{2}}\]
B) \[8\sqrt{5}\,{{(\log 3)}^{2}}\]
C) \[16\sqrt{5}\,(\log 3)\]
D) \[10\sqrt{5}\,{{(\log 3)}^{2}}\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{({{3}^{x}}-1)}^{2}}\,({{3}^{x}}+1)}{\sqrt{5}-\sqrt{4+\cos \,x}}\] \[=2\underset{x\to 0}{\mathop{\lim }}\,\frac{{{({{3}^{x}}-1)}^{2}}\,}{\sqrt{5}-\sqrt{4+\cos \,x}}\] Using L? Hospital?s rule, \[=2\underset{x\to 0}{\mathop{\lim }}\,\frac{2\,\,{{({{3}^{x}}-1)}^{2}}\,\log \,3\,}{-\frac{1}{2}{{(4+\cos x)}^{-1/2}}(-\,\sin x)}\] \[=8\,\log \,3\underset{x\to 0}{\mathop{\lim }}\,\frac{({{3}^{x}}-1)}{{{(4+\cos \,x)}^{-1/2}}\,\sin x}\] \[=8\sqrt{5}\log \,3\,\underset{x\to 0}{\mathop{\lim }}\,\frac{{{3}^{x}}-1}{\sin \,x}\] Again, using L? Hospital?s rule, \[=8\sqrt{5}\log \,3\,\underset{x\to 0}{\mathop{\lim }}\,\frac{{{3}^{x}}-1}{\cos \,x}\] \[=8\sqrt{5}\,\,{{(\log \,3)}^{2}}\]
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