find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the li
$ \lim _ { t \
ightarrow - \infty } \frac { 2 t ^ { 2 } + t } { t ^ { 3 } - 7 t + 1 } $

Question

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by highest power of $t$
Divide numerator $2t^{2}+t$ and denominator $t^{3}-7t + 1$ by $t^{3}$.
$$
\begin{align*}
\frac{2t^{2}+t}{t^{3}-7t + 1}&=\frac{\frac{2t^{2}}{t^{3}}+\frac{t}{t^{3}}}{\frac{t^{3}}{t^{3}}-\frac{7t}{t^{3}}+\frac{1}{t^{3}}}\
&=\frac{\frac{2}{t}+\frac{1}{t^{2}}}{1-\frac{7}{t^{2}}+\frac{1}{t^{3}}}
\end{align*}
$$
## Step2: Apply limit as $t\to-\infty$
We know that $\lim_{t\to-\infty}\frac{1}{t^{n}} = 0$ for $n>0$.
$$
\begin{align*}
\lim_{t\to-\infty}\frac{\frac{2}{t}+\frac{1}{t^{2}}}{1-\frac{7}{t^{2}}+\frac{1}{t^{3}}}&=\frac{\lim_{t\to-\infty}(\frac{2}{t}+\frac{1}{t^{2}})}{\lim_{t\to-\infty}(1-\frac{7}{t^{2}}+\frac{1}{t^{3}})}\
&=\frac{0 + 0}{1-0 + 0}
\end{align*}
$$

# Answer:
$0$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide numerator and denominator by highest power of \(t\)

Step2: Apply limit as \(t\to-\infty\)

Answer: