Question

a function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at negative infinity exists. show that each of the following functions has a horizontal asymptote by calculating the given limit.
$$\lim_{x\to\infty}\frac{-13x}{10 + 2x}=$$
$$\lim_{x\to-\infty}\frac{2x - 15}{x^{3}+2x - 3}=$$
$$\lim_{x\to\infty}\frac{x^{2}-15x - 12}{15 - 10x^{2}}=$$
$$\lim_{x\to\infty}\frac{\sqrt{x^{2}+7x}}{15 - 3x}=$$
$$\lim_{x\to-\infty}\frac{\sqrt{x^{2}+7x}}{15 - 3x}=$$
question help: message instructor

Explanation:

Step1: Calculate \(\lim_{x

ightarrow\infty}\frac{-13x}{10 + 2x}\)
Divide numerator and denominator by \(x\):

$$
LATEXBLOCK0
$$

As \(x
ightarrow\infty\), \(\frac{10}{x}
ightarrow0\). So \(\lim_{x
ightarrow\infty}\frac{-13}{\frac{10}{x}+ 2}=\frac{-13}{0 + 2}=-\frac{13}{2}\)

Step2: Calculate \(\lim_{x

ightarrow-\infty}\frac{2x-15}{x^{3}+2x - 3}\)
Divide numerator and denominator by \(x^{3}\):

$$
LATEXBLOCK1
$$

As \(x
ightarrow-\infty\), \(\frac{2}{x^{2}}
ightarrow0\), \(\frac{15}{x^{3}}
ightarrow0\), \(\frac{2}{x^{2}}
ightarrow0\), \(\frac{3}{x^{3}}
ightarrow0\). So \(\lim_{x
ightarrow-\infty}\frac{\frac{2}{x^{2}}-\frac{15}{x^{3}}}{1+\frac{2}{x^{2}}-\frac{3}{x^{3}}}=\frac{0 - 0}{1+0 - 0}=0\)

Step3: Calculate \(\lim_{x

ightarrow\infty}\frac{x^{2}-15x - 12}{15-10x^{2}}\)
Divide numerator and denominator by \(x^{2}\):

$$
LATEXBLOCK2
$$

As \(x
ightarrow\infty\), \(\frac{15}{x}
ightarrow0\), \(\frac{12}{x^{2}}
ightarrow0\), \(\frac{15}{x^{2}}
ightarrow0\). So \(\lim_{x
ightarrow\infty}\frac{1-\frac{15}{x}-\frac{12}{x^{2}}}{\frac{15}{x^{2}}-10}=\frac{1-0 - 0}{0 - 10}=-\frac{1}{10}\)

Step4: Calculate \(\lim_{x

ightarrow\infty}\frac{\sqrt{x^{2}+7x}}{15-3x}\)
Since \(x
ightarrow\infty\), \(\sqrt{x^{2}+7x}=x\sqrt{1+\frac{7}{x}}\). Divide numerator and denominator by \(x\):

$$
LATEXBLOCK3
$$

As \(x
ightarrow\infty\), \(\frac{7}{x}
ightarrow0\), \(\frac{15}{x}
ightarrow0\). So \(\lim_{x
ightarrow\infty}\frac{\sqrt{1+\frac{7}{x}}}{\frac{15}{x}-3}=\frac{\sqrt{1 + 0}}{0-3}=-\frac{1}{3}\)

Step5: Calculate \(\lim_{x

ightarrow-\infty}\frac{\sqrt{x^{2}+7x}}{15-3x}\)
Since \(x
ightarrow-\infty\), \(\sqrt{x^{2}+7x}=-x\sqrt{1+\frac{7}{x}}\). Divide numerator and denominator by \(x\):

$$
LATEXBLOCK4
$$

As \(x
ightarrow-\infty\), \(\frac{7}{x}
ightarrow0\), \(\frac{15}{x}
ightarrow0\). So \(\lim_{x
ightarrow-\infty}\frac{-\sqrt{1+\frac{7}{x}}}{\frac{15}{x}-3}=\frac{-\sqrt{1+0}}{0 - 3}=\frac{1}{3}\)

Answer:

\(-\frac{13}{2}\)
\(0\)
\(-\frac{1}{10}\)
\(-\frac{1}{3}\)
\(\frac{1}{3}\)