Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to - \infty } \frac{{3{t^2} + t}}{{{t^3} - 4t + 1}} \cr
& {\text{Using properties of limits}} \cr
& \mathop {\lim }\limits_{t \to - \infty } \frac{{3{t^2} + t}}{{{t^3} - 4t + 1}} = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( {3{t^2} + t} \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( {{t^3} - 4t + 1} \right)}} \cr
& {\text{ }} = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( {3{t^2}} \right) + \mathop {\lim }\limits_{t \to - \infty } \left( t \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( {{t^3}} \right) - \mathop {\lim }\limits_{t \to - \infty } \left( {4t} \right) + \mathop {\lim }\limits_{t \to - \infty } \left( 1 \right)}} \cr
& {\text{Evaluate the limits}} \cr
& {\text{ }} = \frac{\infty }{\infty } \cr
& {\text{Divide the numerator and denominator by }}{t^3} \cr
& \mathop {\lim }\limits_{t \to - \infty } \frac{{3{t^2} + t}}{{{t^3} - 4t + 1}} = \mathop {\lim }\limits_{t \to - \infty } \frac{{\frac{{3{t^2}}}{{{t^3}}} + \frac{t}{{{t^3}}}}}{{\frac{{{t^3}}}{{{t^3}}} - \frac{{4t}}{{{t^3}}} + \frac{1}{{{t^3}}}}} \cr
& = \mathop {\lim }\limits_{t \to - \infty } \frac{{\frac{3}{t} + \frac{1}{{{t^2}}}}}{{1 - \frac{4}{{{t^2}}} + \frac{1}{{{t^3}}}}} \cr
& {\text{Use properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{t \to - \infty } \left( {\frac{3}{t}} \right) + \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{{{t^2}}}} \right)}}{{\mathop {\lim }\limits_{t \to - \infty } \left( 1 \right) - \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{4}{{{t^2}}}} \right) + \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{{{t^3}}}} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& = \frac{{0 + 0}}{{1 - 0 + 0}} \cr
& = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{t \to - \infty } \frac{{3{t^2} + t}}{{{t^3} - 4t + 1}} = 0 \cr} $$
