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$\lim\limits_{x\to 0^+}x^m(\log x)^n,(m,n\in N)$ is

$(a)\;0\qquad(b)\;\large\frac{m}{n}\qquad$$(c)\;mn\qquad(d)\;None\;of\;these$

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$\lim\limits_{x\to 0^+}x^m(\log x)^n=\lim\limits_{x\to 0^+}\large\frac{(\log x)^n}{x^{-m}}\qquad\big(\large\frac{0}{0}\big)$
$\lim\limits_{x\to 0^+}\large\frac{n(\log x)^{n-1}.\large\frac{1}{x}}{-mx^{-m-1}}$
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{n(\log x)^{n-1}}{-mx^{-m}}\qquad\big(\large\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{n(n-1)(\log x)^{n-2}.\Large\frac{1}{x}}{(-m^2)x^{-m-1}}$
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{n(n-1)(\log x)^{n-2}}{m^2x^{-m}}\qquad\big(\large\frac{0}{0}\big)$
Continuing like this,after n steps
$\Rightarrow \lim\limits_{x\to 0^+}\large\frac{n!}{(-m)^nx^{-m}}$$=0$
Hence (a) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 
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