A) \[0\]
B) \[\frac{m}{n}\]
C) \[mn\]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{x}^{m}}{{(\log x)}^{n}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{(\log x)}^{n}}}{{{x}^{-m}}},\]\[\left( \frac{\infty }{\infty }Form \right)\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{n{{(\log x)}^{(n-1)}}\frac{1}{x}}{-m{{x}^{-m-1}}}\] [Using L-Hospital's rule] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{n{{(\log x)}^{(n-1)}}}{-m{{x}^{-m}}},\] \[\left( \frac{\infty }{\infty }Form \right)\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{n(n-1){{(\log x)}^{(n-2)}}\frac{1}{x}}{{{(-m)}^{2}}{{x}^{-m-1}}}\] [Again using L-Hospital's rule] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{n(n-1){{(logx)}^{n-2}}}{{{m}^{2}}{{x}^{-m}}},\] \[\left( \frac{\infty }{\infty }Form \right)\] ???????. ???????. \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{n!}{{{(-m)}^{n}}{{x}^{-m}}}=0\]
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