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$\lim _{x \rightarrow 0^{+}}\left(x^{n} \ln x\right), n>0$
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Verified Answer
The correct answer is:
exists and is zero
We have,
$\lim _{x \rightarrow 0^{+}} x^{n} \ln x$
$=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-n}}$
$=\lim _{x \rightarrow 0^{+}} \frac{x}{-n x^{-n-1}} \quad$ [using L'Hospital's rule $]$
$=\lim _{x \rightarrow 0^{+}} \frac{-1}{n x^{-n}}=0 \quad\left[\because \frac{1}{x-n}=0,\right.$ when $\left.x=0\right]$
$\lim _{x \rightarrow 0^{+}} x^{n} \ln x$
$=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-n}}$
$=\lim _{x \rightarrow 0^{+}} \frac{x}{-n x^{-n-1}} \quad$ [using L'Hospital's rule $]$
$=\lim _{x \rightarrow 0^{+}} \frac{-1}{n x^{-n}}=0 \quad\left[\because \frac{1}{x-n}=0,\right.$ when $\left.x=0\right]$
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