Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 61

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$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {x^2}{e^{ - \sqrt x }} \cr & {\text{use the property }}{x^n} = \frac{1}{{{x^{ - n}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - \sqrt x }}}}{{{x^{ - 2}}}} \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 } = {e^{ - \sqrt \infty }} = 0 \cr & \mathop {\lim }\limits_{x \to \infty } {x^{ - 2}} = \frac{1}{{{x^2}}} = \frac{1}{{{{\left( \infty \right)}^2}}} = 0 \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( {{e^{ - \sqrt x }}} \right)}}{{{D_x}\left( {{x^{ - 2}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \frac{1}{{2\sqrt x }}{e^{ - \sqrt x }}}}{{ - 2{x^{ - 3}}}} \cr & = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - \sqrt x }}}}{{{x^{ - 5/2}}}} \cr & {\text{evaluate the limit}} \cr & = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - \infty }}}}{{{{\left( \infty \right)}^{ - 5/2}}}} = \frac{0}{0} \cr & {\text{l'Hospital's rule applies}}{\text{.}} \cr & = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \frac{{{D_x}\left( {{e^{ - \sqrt x }}} \right)}}{{{D_x}\left( {{x^{ - 5/2}}} \right)}} = \frac{1}{4}\mathop {\lim }\limits_{x \to \infty } \frac{{ - \frac{1}{{2\sqrt x }}{e^{ - \sqrt x }}}}{{ - \frac{5}{2}{x^{ - 7/2}}}} \cr & = \frac{1}{{20}}\mathop {\lim }\limits_{x \to \infty } \frac{{{x^{ - 1/2}}{e^{ - \sqrt x }}}}{{{x^{ - 7/2}}}} \cr & {\text{simplifying}} \cr & = \frac{1}{{20}}\mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - \sqrt x }}}}{{{x^3}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{20}}\left( {{{\left( \infty \right)}^{ - 3}}{e^{ - \infty }}} \right) \cr & = 0 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to \infty } {x^2}{e^{ - \sqrt x }} = 0 \cr} $$