Question

evaluate the limit.
\\( \lim _ { t \
ightarrow - \infty } \frac { 2 t + 4 } { 3 t - 6 } \\)
simplify any fractions in your answer.

Explanation:

Step1: Divide numerator and denominator by \(t\)

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

Step2: Evaluate the limit of each term

As \(t
ightarrow -\infty\), \(\lim_{t
ightarrow -\infty}\frac{4}{t}=0\) and \(\lim_{t
ightarrow -\infty}\frac{- 6}{t}=0\).
So \(\lim_{t
ightarrow -\infty}\frac{2+\frac{4}{t}}{3-\frac{6}{t}}=\frac{\lim_{t
ightarrow -\infty}(2)+\lim_{t
ightarrow -\infty}\frac{4}{t}}{\lim_{t
ightarrow -\infty}(3)+\lim_{t
ightarrow -\infty}\frac{-6}{t}}\)
Since \(\lim_{t
ightarrow -\infty}(2) = 2\) and \(\lim_{t
ightarrow -\infty}(3)=3\)

Answer:

\(\frac{2}{3}\)