Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 3

$0.5$

$\int1/x^{3}$ = $-x^{-2}/2$ = $-1/(2x^{2})$ $-1/(2*10^{2})$ - $-1/(2*1^{2})$ = 0.495 $-1/(2*100^{2})$ - $-1/(2*1^{2})$ = 0.49995 $-1/(2*1000^{2})$ - $-1/(2*1^{2})$ = 0.4999995 as x approaches infinity the number gets closer and closer to 0.5, meaning $\lim\limits_{x \to \infty}$ $-1/(2*x^{2})$ - $(-1)/(2*1^{2}) = 0.5$