a function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at…

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.\n$$ lim _ { x \rightarrow infty } \frac { - 4 x } { 15 + 2 x } = $$\n$$ lim _ { x \rightarrow - infty } \frac { 8 x - 5 } { x ^ { 3 } + 12 x - 10 } = $$\n$$ lim _ { x \rightarrow infty } \frac { x ^ { 2 } - 9 x - 2 } { 12 - 13 x ^ { 2 } } = $$\n$$ lim _ { x \rightarrow infty } \frac { sqrt { x ^ { 2 } + 8 x } } { 7 - 5 x } = $$\n$$ lim _ { x \rightarrow - infty } \frac { sqrt { x ^ { 2 } + 8 x } } { 7 - 5 x } = $$

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.\n$$ lim _ { x \rightarrow infty } \frac { - 4 x } { 15 + 2 x } = $$\n$$ lim _ { x \rightarrow - infty } \frac { 8 x - 5 } { x ^ { 3 } + 12 x - 10 } = $$\n$$ lim _ { x \rightarrow infty } \frac { x ^ { 2 } - 9 x - 2 } { 12 - 13 x ^ { 2 } } = $$\n$$ lim _ { x \rightarrow infty } \frac { sqrt { x ^ { 2 } + 8 x } } { 7 - 5 x } = $$\n$$ lim _ { x \rightarrow - infty } \frac { sqrt { x ^ { 2 } + 8 x } } { 7 - 5 x } = $$

Answer

Explanation:

Step1: Solve $\lim_{x\rightarrow\infty}\frac{-4x}{15 + 2x}$

Divide numerator and denominator by (x): $$\lim_{x\rightarrow\infty}\frac{-4x/x}{(15 + 2x)/x}=\lim_{x\rightarrow\infty}\frac{-4}{\frac{15}{x}+2}$$ As (x\rightarrow\infty), (\frac{15}{x}\rightarrow0). So, (\lim_{x\rightarrow\infty}\frac{-4}{\frac{15}{x}+2}=\frac{-4}{0 + 2}=-2)

Step2: Solve $\lim_{x\rightarrow-\infty}\frac{8x - 5}{x^{3}+12x - 10}$

Divide numerator and denominator by (x^{3}): $$\lim_{x\rightarrow-\infty}\frac{(8x - 5)/x^{3}}{(x^{3}+12x - 10)/x^{3}}=\lim_{x\rightarrow-\infty}\frac{\frac{8}{x^{2}}-\frac{5}{x^{3}}}{1+\frac{12}{x^{2}}-\frac{10}{x^{3}}}$$ As (x\rightarrow-\infty), (\frac{8}{x^{2}}\rightarrow0), (\frac{5}{x^{3}}\rightarrow0), (\frac{12}{x^{2}}\rightarrow0), (\frac{10}{x^{3}}\rightarrow0). So, (\lim_{x\rightarrow-\infty}\frac{\frac{8}{x^{2}}-\frac{5}{x^{3}}}{1+\frac{12}{x^{2}}-\frac{10}{x^{3}}}=\frac{0 - 0}{1+0 - 0}=0)

Step3: Solve $\lim_{x\rightarrow\infty}\frac{x^{2}-9x - 2}{12-13x^{2}}$

Divide numerator and denominator by (x^{2}): $$\lim_{x\rightarrow\infty}\frac{(x^{2}-9x - 2)/x^{2}}{(12-13x^{2})/x^{2}}=\lim_{x\rightarrow\infty}\frac{1-\frac{9}{x}-\frac{2}{x^{2}}}{\frac{12}{x^{2}}-13}$$ As (x\rightarrow\infty), (\frac{9}{x}\rightarrow0), (\frac{2}{x^{2}}\rightarrow0), (\frac{12}{x^{2}}\rightarrow0). So, (\lim_{x\rightarrow\infty}\frac{1-\frac{9}{x}-\frac{2}{x^{2}}}{\frac{12}{x^{2}}-13}=\frac{1-0 - 0}{0 - 13}=-\frac{1}{13})

Step4: Solve $\lim_{x\rightarrow\infty}\frac{\sqrt{x^{2}+8x}}{7-5x}$

Since (x\rightarrow\infty), (\sqrt{x^{2}+8x}=x\sqrt{1+\frac{8}{x}}). Divide numerator and denominator by (x): $$\lim_{x\rightarrow\infty}\frac{\sqrt{x^{2}+8x}/x}{(7 - 5x)/x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{8}{x}}}{\frac{7}{x}-5}$$ As (x\rightarrow\infty), (\frac{8}{x}\rightarrow0), (\frac{7}{x}\rightarrow0). So, (\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{8}{x}}}{\frac{7}{x}-5}=\frac{\sqrt{1+0}}{0 - 5}=-\frac{1}{5})

Step5: Solve $\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+8x}}{7-5x}$

Since (x\rightarrow-\infty), (\sqrt{x^{2}+8x}=-x\sqrt{1+\frac{8}{x}}). Divide numerator and denominator by (x): $$\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+8x}/x}{(7 - 5x)/x}=\lim_{x\rightarrow-\infty}\frac{-\sqrt{1+\frac{8}{x}}}{\frac{7}{x}-5}$$ As (x\rightarrow-\infty), (\frac{8}{x}\rightarrow0), (\frac{7}{x}\rightarrow0). So, (\lim_{x\rightarrow-\infty}\frac{-\sqrt{1+\frac{8}{x}}}{\frac{7}{x}-5}=\frac{-\sqrt{1+0}}{0 - 5}=\frac{1}{5})

Answer:

-2, 0, (-\frac{1}{13}), (-\frac{1}{5}), (\frac{1}{5})