$\lim _{t \rightarrow-\infty} \frac{3 t^{2}+t}{t^{3}-4 t+1}$

Step 1 ：We are given the limit: \(\lim _{t \rightarrow-\infty} \frac{3 t^{2}+t}{t^{3}-4 t+1}\)

Step 2 ：This limit is of the form \(\frac{\infty}{\infty}\), which is an indeterminate form. We can use L'Hopital's rule to solve this.

Step 3 ：L'Hopital's rule states that if the limit of a function is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit of that function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Step 4 ：Let's find the derivatives of the numerator and the denominator. The derivative of the numerator \(3t^{2} + t\) is \(6t + 1\) and the derivative of the denominator \(t^{3} - 4t + 1\) is \(3t^{2} - 4\).

Step 5 ：Applying L'Hopital's rule, we get the new limit: \(\lim _{t \rightarrow-\infty} \frac{6t + 1}{3t^{2} - 4}\)

Step 6 ：This limit is also of the form \(\frac{\infty}{\infty}\), so we can apply L'Hopital's rule again. The derivative of the numerator \(6t + 1\) is \(6\) and the derivative of the denominator \(3t^{2} - 4\) is \(6t\).

Step 7 ：Applying L'Hopital's rule again, we get the new limit: \(\lim _{t \rightarrow-\infty} \frac{6}{6t}\)

Step 8 ：This limit is not of the form \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\), so we can directly compute it. As \(t\) approaches negative infinity, the fraction approaches 0.

Step 9 ：Final Answer: \(\boxed{0}\)