Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises - Page 96: 27

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 2{\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = 2 \cr & {\text{horizontal asymptote}}\,\,y = 2 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{6{x^2} - 9x + 8}}{{3{x^2} + 2}} \cr & {\text{Determine }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{6{x^2} - 9x + 8}}{{3{x^2} + 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {6{x^2} - 9x + 8} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {3{x^2} + 2} \right)}} \cr & {\text{Use property of limits}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6\mathop {\lim }\limits_{x \to \infty } \left( {{x^2}} \right) - 9\mathop {\lim }\limits_{x \to \infty } \left( x \right) + \mathop {\lim }\limits_{x \to \infty } \left( 8 \right)}}{{3\mathop {\lim }\limits_{x \to \infty } \left( {{x^2}} \right) + 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\infty }{\infty } \cr & {\text{We divide the numerator and denominator by the largest }} \cr & {\text{power of }}x{\text{ appearing in the denominator}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{6{x^2} - 9x + 8}}{{3{x^2} + 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{6{x^2}}}{{{x^2}}} - \frac{{9x}}{{{x^2}}} + \frac{8}{{{x^2}}}}}{{\frac{{3{x^2}}}{{{x^2}}} + \frac{2}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{6 - \frac{9}{x} + \frac{8}{{{x^2}}}}}{{3 + \frac{2}{{{x^2}}}}} \cr & \,\, = \mathop {\lim }\limits_{x \to \infty } \frac{{\mathop {\lim }\limits_{x \to \infty } 6 - \overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {\frac{9}{x}} \right)}^{{\text{approaches0}}} + \overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {\frac{8}{{{x^2}}}} \right)}^{{\text{approaches0}}}}}{{3 + \underbrace {\mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{{{x^2}}}} \right)}_{{\text{approaches0}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6 - 0 + 0}}{{3 + 0}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cr & \cr & {\text{Determine }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr & {\text{Using the same steps}},{\text{ it can be shown that}} \cr & \mathop {\lim }\limits_{x \to - \infty } \frac{{6{x^2} - 9x + 8}}{{3{x^2} + 2}} = 2 \cr & \cr & {\text{Then, the horizontal asymptote of }}f\left( x \right){\text{ is}} \cr & y = 2 \cr} $$