Question

evaluate the limit.

$\lim_{t\to\infty}\frac{-6}{-4t^{3}-3t^{2}-2t - 3}$

simplify any fractions in your answer.

Explanation:

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

$$\lim_{t ightarrow\infty}\frac{\frac{- 6}{t^{3}}}{\frac{-4t^{3}}{t^{3}}+\frac{-3t^{2}}{t^{3}}+\frac{-2t}{t^{3}}+\frac{-3}{t^{3}}}$$

Step2: Use the limit property \(\lim_{t

ightarrow\infty}\frac{1}{t^{n}} = 0\) (\(n>0\))

$$\lim_{t ightarrow\infty}\frac{\frac{-6}{t^{3}}}{-4-\frac{3}{t}-\frac{2}{t^{2}}-\frac{3}{t^{3}}}$$

Since \(\lim_{t
ightarrow\infty}\frac{-6}{t^{3}}=0\), \(\lim_{t
ightarrow\infty}\frac{3}{t}=0\), \(\lim_{t
ightarrow\infty}\frac{2}{t^{2}} = 0\), \(\lim_{t
ightarrow\infty}\frac{3}{t^{3}}=0\)

Step3: Calculate the limit

$$\frac{0}{-4 - 0-0 - 0}=0$$

Answer:

\(0\)