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. $lim_{x
ightarrowinfty}\frac{-5x}{5 + 2x}=square$ $lim_{x
ightarrow-infty}\frac{14x - 10}{x^{2}+12x - 6}=square$ $lim_{x
ightarrowinfty}\frac{14x - 10}{x^{2}+12x - 6}=square$ $lim_{x
ightarrow-infty}\frac{sqrt{8 - 4x^{2}}}{sqrt{x^{2}+11x}}=square$ $lim_{x
ightarrowinfty}\frac{sqrt{8 - 4x^{2}}}{sqrt{x^{2}+11x}}=square$ $lim_{x
ightarrow-infty}\frac{2 - 7x}{sqrt{x^{2}+11x}}=square$ $lim_{x
ightarrowinfty}\frac{2 - 7x}{sqrt{x^{2}+11x}}=square$

Explanation:

Step1: Divide numerator and denominator by highest - power of x in denominator for $\lim_{x

ightarrow\infty}\frac{-5x}{5 + 2x}$
Divide numerator and denominator by $x$: $\lim_{x
ightarrow\infty}\frac{-5x/x}{(5 + 2x)/x}=\lim_{x
ightarrow\infty}\frac{-5}{\frac{5}{x}+2}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{5}{x}=0$. So, $\lim_{x
ightarrow\infty}\frac{-5}{\frac{5}{x}+2}=\frac{-5}{0 + 2}=-\frac{5}{2}$.

Step2: Divide numerator and denominator by highest - power of x in denominator for $\lim_{x

ightarrow-\infty}\frac{14x-10}{x^{2}+12x - 6}$
Divide numerator and denominator by $x^{2}$: $\lim_{x
ightarrow-\infty}\frac{\frac{14x}{x^{2}}-\frac{10}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{12x}{x^{2}}-\frac{6}{x^{2}}}=\lim_{x
ightarrow-\infty}\frac{\frac{14}{x}-\frac{10}{x^{2}}}{1+\frac{12}{x}-\frac{6}{x^{2}}}$.
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{14}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{10}{x^{2}} = 0$, $\lim_{x
ightarrow-\infty}\frac{12}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{6}{x^{2}}=0$. So, $\lim_{x
ightarrow-\infty}\frac{\frac{14}{x}-\frac{10}{x^{2}}}{1+\frac{12}{x}-\frac{6}{x^{2}}}=0$.

Step3: Divide numerator and denominator by highest - power of x in denominator for $\lim_{x

ightarrow\infty}\frac{x^{2}-13x - 11}{8 - 4x^{2}}$
Divide numerator and denominator by $x^{2}$: $\lim_{x
ightarrow\infty}\frac{\frac{x^{2}}{x^{2}}-\frac{13x}{x^{2}}-\frac{11}{x^{2}}}{\frac{8}{x^{2}}-\frac{4x^{2}}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{1-\frac{13}{x}-\frac{11}{x^{2}}}{\frac{8}{x^{2}}-4}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{13}{x}=0$, $\lim_{x
ightarrow\infty}\frac{11}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{8}{x^{2}}=0$. So, $\lim_{x
ightarrow\infty}\frac{1-\frac{13}{x}-\frac{11}{x^{2}}}{\frac{8}{x^{2}}-4}=-\frac{1}{4}$.

Step4: Analyze $\lim_{x

ightarrow\infty}\frac{\sqrt{8 - 4x^{2}}}{\sqrt{x^{2}+11x}}$
For large positive $x$, $\sqrt{8 - 4x^{2}}\approx\sqrt{-4x^{2}}$ (dominant term) and $\sqrt{x^{2}+11x}\approx\sqrt{x^{2}}=x$. But we rewrite as $\lim_{x
ightarrow\infty}\frac{\sqrt{x^{2}(\frac{8}{x^{2}}-4)}}{\sqrt{x^{2}(1 + \frac{11}{x})}}=\lim_{x
ightarrow\infty}\frac{|x|\sqrt{\frac{8}{x^{2}}-4}}{|x|\sqrt{1+\frac{11}{x}}}$. Since $x
ightarrow\infty$, $|x| = x$ and $\lim_{x
ightarrow\infty}\frac{\sqrt{\frac{8}{x^{2}}-4}}{\sqrt{1+\frac{11}{x}}}=\sqrt{\frac{0 - 4}{1+0}}=- 2$.

Step5: Analyze $\lim_{x

ightarrow-\infty}\frac{2-7x}{\sqrt{x^{2}+11x}}$
For $x
ightarrow-\infty$, $|x|=-x$. $\lim_{x
ightarrow-\infty}\frac{2-7x}{\sqrt{x^{2}(1+\frac{11}{x})}}=\lim_{x
ightarrow-\infty}\frac{2-7x}{|x|\sqrt{1+\frac{11}{x}}}=\lim_{x
ightarrow-\infty}\frac{2-7x}{-x\sqrt{1+\frac{11}{x}}}=\lim_{x
ightarrow-\infty}\frac{\frac{2}{x}-7}{-\sqrt{1+\frac{11}{x}}}=7$.

Step6: Analyze $\lim_{x

ightarrow\infty}\frac{\sqrt{x^{2}+11x}}{2-7x}$
For $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{\sqrt{x^{2}(1+\frac{11}{x})}}{2-7x}=\lim_{x
ightarrow\infty}\frac{|x|\sqrt{1+\frac{11}{x}}}{2-7x}=\lim_{x
ightarrow\infty}\frac{x\sqrt{1+\frac{11}{x}}}{2-7x}=\lim_{x
ightarrow\infty}\frac{\sqrt{1+\frac{11}{x}}}{\frac{2}{x}-7}=-\frac{1}{7}$.

Answer:

$-\frac{5}{2},0,-\frac{1}{4},-2,7,-\frac{1}{7}$