Question

find the horizontal asymptote
f(x) = \frac{3x(x + 2)}{x^{2}-1}

Explanation:

Step1: Expand the numerator

Expand \(3x(x + 2)\) to get \(3x^{2}+6x\). So the function becomes \(f(x)=\frac{3x^{2}+6x}{x^{2}-1}\).

Step2: Divide numerator and denominator by \(x^{2}\)

$$
LATEXBLOCK0
$$

Step3: Evaluate the limit

As \(x
ightarrow\pm\infty\), \(\frac{6}{x}
ightarrow0\) and \(\frac{1}{x^{2}}
ightarrow0\). So \(\lim_{x
ightarrow\pm\infty}\frac{3+\frac{6}{x}}{1-\frac{1}{x^{2}}}=\frac{3 + 0}{1-0}=3\)

Answer:

The horizontal asymptote is \(y = 3\)