A) 0
B) \[\frac{m}{n}\]
C) \[mn\]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0+}{\mathop{\lim }}\,\,{{x}^{m}}\,{{(\log x)}^{n}}=\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{{{(\log x)}^{n}}}{{{x}^{-m}}}\] \[\left( \text{Form}\frac{\infty }{\infty } \right)\] \[\log \,\,P=\int_{0}^{1}{{}}\log x\,dx=(x\,\log x-x)_{0}^{1}=(-1)\] (By L-Hospital's rule) \[=\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{n\,{{(\log x)}^{n-1}}}{-m{{x}^{-m}}}\,\] \[\left( \text{Form}\frac{\infty }{\infty } \right)\] \[=\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{n\,(n-1)\,{{(\log x)}^{(n-2)}}\frac{1}{x}}{{{(-m)}^{2}}{{x}^{-m-1}}}\] (By L-Hospital's rule) \[=\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{n\,(n-1)\,{{(\log x)}^{n-2}}}{{{m}^{2}}{{x}^{-m}}}\,\] \[\left( \text{Form}\frac{\infty }{\infty } \right)\] ....................... ...................... \[=\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{n\,\,!}{{{(-m)}^{n}}{{x}^{-m}}}=0\] (Differentiating \[{{N}^{r}}\] and \[{{D}^{r}}\] n times).
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