The value of mathop {lim }limits_{x to {0^ + | vedclass

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Question

(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0^ + } x^m (\log x)^n$ as $x 	o 0^+$.This is an indeterminate form of type $0 	imes \infty$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0^ + } x^m (\log x)^n$ as $x 	o 0^+$.This is an indeterminate form of type $0 	imes \infty$.We rewrite the expression as $\mathop {\lim }\limits_{x 	o 0^ + } \frac{(\log x)^n}{x^{-m}}$,which is of the form $\frac{\infty}{\infty}$.Applying $L'	ext{Hospital's rule}$ $n$ times,we differentiate the numerator and denominator with respect to $x$ repeatedly.The $k$-th derivative of $(\log x)^n$ involves terms of $(\log x)^{n-k}$.After $n$ applications of $L'	ext{Hospital's rule}$,the numerator becomes a constant $n!$ and the denominator remains a power of $x$ (specifically $x^{-m}$ multiplied by constants).Specifically,$\mathop {\lim }\limits_{x 	o 0^ + } \frac{n! \cdot (-1)^n}{(-m)^n x^{-m}} = \mathop {\lim }\limits_{x 	o 0^ + } \frac{n! \cdot (-1)^n x^m}{(-m)^n}$.Since $m > 0$,as $x 	o 0^+$,$x^m 	o 0$.Therefore,the limit is $0$.

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

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