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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is

$\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$ is equal to

$\lim _{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module