Question

what is the horizontal asymptote of $f(x)=\frac{3x}{x + 2}$?

Explanation:

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

$$\lim_{x ightarrow\pm\infty}\frac{3x}{x + 2}=\lim_{x ightarrow\pm\infty}\frac{3}{1+\frac{2}{x}}$$

Step2: Evaluate the limit

As \(x
ightarrow\pm\infty\), \(\frac{2}{x}
ightarrow0\). So, \(\lim_{x
ightarrow\pm\infty}\frac{3}{1+\frac{2}{x}}=\frac{3}{1 + 0}=3\)

Answer:

\(y = 3\)