Answer
$$\left( {\bf{a}} \right)\mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{{{x^n}}} = 0{\text{ and }}\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}}}{{\ln x}} = + \infty {\text{ Proved}}$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{{{x^n}}} = 0,{\text{ }}n{\text{ is any positive integer}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{{{x^n}}} = \frac{{\ln \left( { + \infty } \right)}}{{{{\left( { + \infty } \right)}^n}}} = \frac{\infty }{\infty } \cr
& {\text{Apply the L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{{{x^n}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{d}{{dx}}\left[ {\ln x} \right]}}{{\frac{d}{{dx}}\left[ {{x^n}} \right]}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x}}}{{n{x^{n - 1}}}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x}}}{{n{x^{n - 1}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{n{x^n}}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{n{x^n}}} = \frac{1}{{n{{\left( { + \infty } \right)}^n}}} = \frac{1}{\infty } = 0 \cr
& \cr
& \left( {\text{b}} \right)\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}}}{{\ln x}} = + \infty \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}}}{{\ln x}} = \frac{{{{\left( { + \infty } \right)}^n}}}{{\ln \left( { + \infty } \right)}} = \frac{\infty }{\infty } \cr
& {\text{Apply the L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}}}{{\ln x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{d}{{dx}}\left[ {{x^n}} \right]}}{{\frac{d}{{dx}}\left[ {\ln x} \right]}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{n{x^{n - 1}}}}{{\frac{1}{x}}} \cr
& \frac{{{x^n}}}{{\ln x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{n{x^n}}}{1} = \mathop {\lim }\limits_{x \to + \infty } n{x^n} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to + \infty } n{x^n} = + \infty \cr} $$
