$Find $\lim _{x \rightarrow 0^{+}} \sqrt{x}\ln x$$

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How should I compute this limit?

Calculus Limits Logarithms

Tpolo

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You can use the L'hopital's rule. We have
\[\lim_{x \rightarrow 0} \sqrt{x}\ln x=\lim_{x \rightarrow 0} x^{\frac{1}{2}}\ln x=\lim_{x \rightarrow 0} \frac{\ln x}{x^{-\frac{1}{2}}}=\lim_{x \rightarrow 0} \frac{(\ln x)'}{(x^{-\frac{1}{2}})'}\]
\[=\lim_{x \rightarrow 0} \frac{\frac{1}{x}}{-\frac{1}{2}x^{-\frac{3}{2}}}=\lim_{x \rightarrow 0} -2 x^{\frac{3}{2}}\frac{1}{x}=\lim_{x \rightarrow 0}-2 x^{\frac{1}{2}}=0.\]
Hence
\[\lim_{x \rightarrow 0} \sqrt{x}\ln x=0.\]

Nirenberg

$Find $\lim _{x \rightarrow 0^{+}} \sqrt{x}\ln x$$

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