Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 62

$${\text{The limit does not exist}}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{\ln \left( {{x^3} + 1} \right)}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } = \ln \left( {\sqrt \infty } \right) = \infty \cr & \mathop {\lim }\limits_{x \to \infty } \ln \left( {{\infty ^3} + 1} \right) = \infty \cr & {\text{Since the limits of both numerator and denominator are }}\infty {\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr & {\text{denominator}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{D_x}\left( {\sqrt x } \right)}}{{{D_x}\left( {\ln \left( {{x^3} + 1} \right)} \right)}} \cr & {D_x}\left( {\sqrt x } \right) = \frac{1}{{2\sqrt x }} \cr & {D_x}\left( {\ln \left( {{x^3} + 1} \right)} \right) = \frac{{3{x^2}}}{{{x^3} + 1}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{2\sqrt x }}}}{{\frac{{3{x^2}}}{{{x^3} + 1}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 1}}{{3{x^{5/2}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^3}}}{{3{x^{5/2}}}} + \frac{1}{{3{x^{5/2}}}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{3}{x^{1/2}} + \frac{1}{{3{x^{5/2}}}}} \right) \cr & {\text{Find the limit }} \cr & = \frac{1}{3}{\left( \infty \right)^{1/2}} + \frac{1}{{3{{\left( \infty \right)}^{5/2}}}} \cr & = \frac{1}{3}{\left( \infty \right)^{1/2}} + 0 \cr & = \infty \cr & {\text{The limit does not exist}} \cr} $$