Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise exist, enter dne.)
lim_{t \to -infty} \frac{2 t^{2}+t}{t^{3}-7 t+1}
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Explanation:

Step1: Divide numerator and denominator by \(t^{3}\)

$$\lim_{t ightarrow-\infty}\frac{2t^{2}+t}{t^{3}-7t + 1}=\lim_{t ightarrow-\infty}\frac{\frac{2t^{2}}{t^{3}}+\frac{t}{t^{3}}}{\frac{t^{3}}{t^{3}}-\frac{7t}{t^{3}}+\frac{1}{t^{3}}}$$

Simplify the fractions:

$$=\lim_{t ightarrow-\infty}\frac{\frac{2}{t}+\frac{1}{t^{2}}}{1-\frac{7}{t^{2}}+\frac{1}{t^{3}}}$$

Step2: Evaluate the limit

As \(t
ightarrow-\infty\), \(\frac{2}{t}
ightarrow0\), \(\frac{1}{t^{2}}
ightarrow0\), \(\frac{7}{t^{2}}
ightarrow0\) and \(\frac{1}{t^{3}}
ightarrow0\)

$$\lim_{t ightarrow-\infty}\frac{\frac{2}{t}+\frac{1}{t^{2}}}{1-\frac{7}{t^{2}}+\frac{1}{t^{3}}}=\frac{0 + 0}{1-0 + 0}$$

Answer:

\(0\)