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The value of $\lim _{x \rightarrow 0^{+}} x^m(\log x)^n, m, n \in N$ is
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$0$
$\lim _{x \rightarrow 0+} x^m(\log x)^n=\lim _{x \rightarrow 0+} \frac{(\log x)^n}{x^{-m}}\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{(n-1)} \frac{1}{x}}{-m x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{n-1}}{-m x^{-m}}$
$\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{(n-2)} \frac{1}{x}}{(-m)^2 x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{n-2}}{m^2 x^{-m}} \quad\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
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$=\lim _{x \rightarrow 0+} \frac{n!}{(-m)^n x^{-m}}=0$
(Differentiating $N^r$ and $D^r n$ times).
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{(n-1)} \frac{1}{x}}{-m x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{n-1}}{-m x^{-m}}$
$\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{(n-2)} \frac{1}{x}}{(-m)^2 x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{n-2}}{m^2 x^{-m}} \quad\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
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$=\lim _{x \rightarrow 0+} \frac{n!}{(-m)^n x^{-m}}=0$
(Differentiating $N^r$ and $D^r n$ times).
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