Question

question
find all horizontal asymptotes of the following function.

f(x)=\frac{15x - 9}{5x + 9}

Explanation:

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

$$\begin{align*} \lim_{x ightarrow\infty}\frac{15x - 9}{5x+9}&=\lim_{x ightarrow\infty}\frac{\frac{15x}{x}-\frac{9}{x}}{\frac{5x}{x}+\frac{9}{x}}\\ &=\lim_{x ightarrow\infty}\frac{15-\frac{9}{x}}{5 + \frac{9}{x}} \end{align*}$$

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

ightarrow\infty}\frac{c}{x}=0\) (\(c\) is a constant)
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{9}{x}=0\). Then \(\lim_{x
ightarrow\infty}\frac{15-\frac{9}{x}}{5+\frac{9}{x}}=\frac{15 - 0}{5+0}=3\)

Step3: Check the limit as \(x

ightarrow-\infty\)

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{15x - 9}{5x+9}&=\lim_{x ightarrow-\infty}\frac{\frac{15x}{x}-\frac{9}{x}}{\frac{5x}{x}+\frac{9}{x}}\\ &=\lim_{x ightarrow-\infty}\frac{15-\frac{9}{x}}{5+\frac{9}{x}} \end{align*}$$

As \(x
ightarrow-\infty\), \(\lim_{x
ightarrow-\infty}\frac{9}{x}=0\). So \(\lim_{x
ightarrow-\infty}\frac{15-\frac{9}{x}}{5+\frac{9}{x}}=\frac{15-0}{5 + 0}=3\)

Answer:

The horizontal asymptote is \(y = 3\)