Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 138: 18

$- 3$

$$\eqalign{ & \mathop {\lim }\limits_{t \to - \infty } \frac{{6{t^2} + t - 5}}{{9 - 2{t^2}}} \cr & {\text{Using properties of limits}} \cr & \mathop {\lim }\limits_{t \to - \infty } \frac{{6{t^2} + t - 5}}{{9 - 2{t^2}}} = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( {6{t^2} + t - 5} \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( {9 - 2{t^2}} \right)}} \cr & {\text{ }} = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( {6{t^2}} \right) + \mathop {\lim }\limits_{t \to - \infty } \left( t \right) - \mathop {\lim }\limits_{t \to - \infty } \left( 5 \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( 9 \right) - \mathop {\lim }\limits_{t \to - \infty } \left( {2{t^2}} \right)}} \cr & {\text{Evaluate the limits}} \cr & {\text{ }} = \frac{\infty }{\infty } \cr & {\text{Divide the numerator and denominator by }}{t^2} \cr & \mathop {\lim }\limits_{t \to - \infty } \frac{{6{t^2} + t - 5}}{{9 - 2{t^2}}} = \mathop {\lim }\limits_{t \to - \infty } \frac{{\frac{{6{t^2}}}{{{t^2}}} + \frac{t}{{{t^2}}} - \frac{5}{{{t^2}}}}}{{\frac{9}{{{t^2}}} - \frac{{2{t^2}}}{{{t^2}}}}} \cr & = \mathop {\lim }\limits_{t \to - \infty } \frac{{6 + \frac{1}{t} - \frac{5}{{{t^2}}}}}{{\frac{9}{{{t^2}}} - 2}} \cr & {\text{Use properties of limits}} \cr & = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( 6 \right) + \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{t}} \right) - \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{5}{{{t^2}}}} \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( {\frac{9}{{{t^2}}}} \right) - \mathop {\lim }\limits_{t \to - \infty } \left( 2 \right)}} \cr & {\text{Evaluate the limits}} \cr & = \frac{{6 + 0 - 0}}{{0 - 2}} \cr & = - 3 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{t \to - \infty } \frac{{6{t^2} + t - 5}}{{9 - 2{t^2}}} = - 3 \cr} $$