Question

find the horizontal asymptote.
$y = \frac{3x - 3}{x + 1}$
$y = ?$

Explanation:

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

$$y=\lim_{x ightarrow\pm\infty}\frac{3x - 3}{x + 1}=\lim_{x ightarrow\pm\infty}\frac{\frac{3x}{x}-\frac{3}{x}}{\frac{x}{x}+\frac{1}{x}}$$

Step2: Evaluate the limit

As \(x
ightarrow\pm\infty\), \(\frac{3}{x}
ightarrow0\) and \(\frac{1}{x}
ightarrow0\). So \(y = \frac{3-0}{1 + 0}\)

Answer:

\(y = 3\)