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

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

Question

(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 $0$

Vedclass · Answer

(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 $n > 0$.

Similar Questions

Evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

$\mathop {\lim }\limits_{x \to a} \frac{{\cos x - \cos a}}{{\cot x - \cot a}} = $

$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$ is equal to

If $f$ is a strictly increasing function,then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to

The $\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module