A) \[\log n\left( \frac{2}{3} \right)\]
B) 0
C) \[n\log n\left( \frac{2}{3} \right)\]
D) Not defined
Correct Answer: B
Solution :
[b] \[L=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{({{2}^{{{x}^{n}}}})}^{\frac{1}{{{e}^{x}}}}}-{{({{3}^{{{x}^{n}}}})}^{\frac{1}{{{e}^{x}}}}}}{{{x}^{n}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{(3){{\,}^{^{\frac{{{x}^{n}}}{{{e}^{x}}}}}}\left( {{\left( \frac{2}{3} \right)}^{\frac{{{x}^{n}}}{{{e}^{x}}}}}-1 \right)}{{{x}^{n}}}\] Now, \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{n}}}{{{e}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{n!}{{{e}^{x}}}=0\] (Applying L?Hospital?s rule n times) Hence, \[L=\underset{x\to \infty }{\mathop{\lim }}\,{{(3)}^{\frac{{{x}^{n}}}{{{e}^{x}}}}}\underset{x\to \infty }{\mathop{\lim }}\,\frac{\left( {{\left( \frac{2}{3} \right)}^{\frac{{{x}^{n}}}{{{e}^{x}}}}}-1 \right)}{\frac{{{x}^{n}}}{{{e}^{x}}}}\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{{{e}^{x}}}\] \[=1\times \log (2/3)\times 0=0.\]
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