Evaluate lim _{x rightarrow 0^{+}} (x^{n} ln | vedclass

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(B) We have $\lim _{x ightarrow 0^{+}} x^{n} \ln x$. This is an indeterminate form of type $0 	imes \infty$. We can rewrite it as $\lim _{x ighta

Vedclass · Accepted Answer

exists and is $0$

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(B) We have $\lim _{x ightarrow 0^{+}} x^{n} \ln x$. This is an indeterminate form of type $0 	imes \infty$. We can rewrite it as $\lim _{x ightarrow 0^{+}} \frac{\ln x}{x^{-n}}$. Applying $L'	ext{Hospital's rule}$, we differentiate the numerator and denominator with respect to $x$: $= \lim _{x ightarrow 0^{+}} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^{-n})} = \lim _{x ightarrow 0^{+}} \frac{\frac{1}{x}}{-n x^{-n-1}}$. $= \lim _{x ightarrow 0^{+}} \frac{1}{x} \cdot \frac{x^{n+1}}{-n} = \lim _{x ightarrow 0^{+}} \frac{x^{n}}{-n}$. Since $n > 0$, as $x ightarrow 0^{+}$, $x^{n} ightarrow 0$. Therefore, the limit is $\frac{0}{-n} = 0$.

Evaluate $\lim _{x \rightarrow 0^{+}} (x^{n} \ln x)$ for $n > 0$.

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