Let l = mathop {Lim}limits_{x to {0^ + }} x^m | vedclass

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Question

(A) We are given the limit $l = \mathop {Lim}\limits_{x 	o {0^ + }} x^m (\ln x)^n$.Rewrite the expression as $l = \mathop {Lim}\limits_{x 	o {0^ + }

Vedclass · Accepted Answer

$l$ is independent of $m$ and $n$

Vedclass · Answer

(A) We are given the limit $l = \mathop {Lim}\limits_{x 	o {0^ + }} x^m (\ln x)^n$.Rewrite the expression as $l = \mathop {Lim}\limits_{x 	o {0^ + }} \frac{(\ln x)^n}{x^{-m}}$.Since this is an indeterminate form of type $\frac{\infty}{\infty}$ as $x 	o 0^+$,we apply $L'	ext{Hospital's rule}$ $n$ times.After applying the rule $n$ times,the numerator becomes $n!$ (multiplied by $(-1)^n$) and the denominator involves $x^m$ multiplied by constants.Specifically,$l = \mathop {Lim}\limits_{x 	o {0^ + }} \frac{n! (-1)^n}{(-m)^n x^{-m}} = \mathop {Lim}\limits_{x 	o {0^ + }} \frac{n! (-1)^n x^m}{(-m)^n} = 0$.Since the limit $l = 0$ regardless of the specific natural numbers $m$ and $n$,the value of $l$ is independent of both $m$ and $n$.

Let $l = \mathop {Lim}\limits_{x \to {0^ + }} x^m (\ln x)^n$ where $m, n \in N$,then:

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