The value of mathop {lim }limits_{x to infty | vedclass

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(A) Given the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{\log x}{x^n}$ where $n > 0$.Since the form is $\frac{\infty}{\infty}$ as $x 	o \in

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$0$

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(A) Given the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{\log x}{x^n}$ where $n > 0$.Since the form is $\frac{\infty}{\infty}$ as $x 	o \infty$,we apply $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o \infty } \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x^n)} = \mathop {\lim }\limits_{x 	o \infty } \frac{\frac{1}{x}}{n x^{n-1}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{1}{n x^n}$As $x 	o \infty$,$x^n 	o \infty$ for $n > 0$,so the expression approaches $0$.

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{{x^n}}}, \; n > 0$ is

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