find all horizontal asymptotes of the following function. f(x)=(3x + 4)(x + 5)/3(3x + 10)(x + 2) answer one…

Answer:

$y = \frac{1}{3}$

Explanation:

Step1: Expand the numerator

$(3x + 4)(x + 5)=3x^{2}+15x+4x + 20=3x^{2}+19x + 20$

Step2: Expand the denominator

$3(3x + 10)(x + 2)=3(3x^{2}+6x+10x + 20)=9x^{2}+48x + 60$

Step3: Find the limit as $x\to\pm\infty$

$\lim_{x\to\pm\infty}\frac{3x^{2}+19x + 20}{9x^{2}+48x + 60}$. Divide numerator and denominator by $x^{2}$: $\lim_{x\to\pm\infty}\frac{3+\frac{19}{x}+\frac{20}{x^{2}}}{9+\frac{48}{x}+\frac{60}{x^{2}}}$. As $x\to\pm\infty$, $\frac{19}{x}\to0$, $\frac{20}{x^{2}}\to0$, $\frac{48}{x}\to0$, $\frac{60}{x^{2}}\to0$. So the limit is $\frac{3}{9}=\frac{1}{3}$.