Question

evaluate the limit.

lim_{t\to -\infty}\frac{9t^{2}+10t + 7}{-7t + 7}

simplify any fractions in your answer.

Explanation:

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

The highest power of \(t\) in the denominator \(-7t + 7\) is \(t\).
Divide each term in \(\frac{9t^{2}+10t + 7}{-7t + 7}\) by \(t\):

$$ \frac{\frac{9t^{2}}{t}+\frac{10t}{t}+\frac{7}{t}}{\frac{-7t}{t}+\frac{7}{t}}=\frac{9t + 10+\frac{7}{t}}{-7+\frac{7}{t}} $$

Step2: Apply the limit \(\lim_{t

ightarrow-\infty}\)
We know that \(\lim_{t
ightarrow-\infty}\frac{1}{t}=0\).

$$ \lim_{t ightarrow-\infty}\frac{9t + 10+\frac{7}{t}}{-7+\frac{7}{t}}=\frac{\lim_{t ightarrow-\infty}(9t)+\lim_{t ightarrow-\infty}10+\lim_{t ightarrow-\infty}\frac{7}{t}}{\lim_{t ightarrow-\infty}(-7)+\lim_{t ightarrow-\infty}\frac{7}{t}} $$

Since \(\lim_{t
ightarrow-\infty}\frac{7}{t}=0\) and \(\lim_{t
ightarrow-\infty}10 = 10\), \(\lim_{t
ightarrow-\infty}(-7)=-7\), and \(\lim_{t
ightarrow-\infty}(9t)=-\infty\)

$$ \frac{-\infty+10 + 0}{-7+0}=-\infty $$

Answer:

\(-\infty\)