## Calculus: Early Transcendentals 8th Edition

Vertical: $x = -3$ Horizontal: $y = 4$
To find the horizontal asymptote, we need to find: $\lim\limits_{x \to \infty} \frac{5 + 4x}{x+3}$ and $\lim\limits_{x \to -\infty} \frac{5 + 4x}{x+3}$ First we divide both numerator and denominator by $x$. $\lim\limits_{x \to \infty} \frac{\frac{5+4x}{x}}{\frac{x+3}{x}} =$ $\lim\limits_{x \to \infty} \frac{\frac{5}{x}+4}{\frac{3}{x}+1} = \frac{0+4}{1+0} = 4$ Now with $-\infty$. $\lim\limits_{x \to -\infty} \frac{\frac{5+4x}{x}}{\frac{x+3}{x}} =$ $\lim\limits_{x \to -\infty} \frac{\frac{5}{x}+4}{\frac{3}{x}+1} = \frac{0+4}{1+0} = 4$ So with both answers we know that the horizontal asymptote is $y = 4$. To find the vertical asymptote we take the denominator $x+3$ and compare it to zero. $x + 3 = 0$ $x = -3$ So the vertical asymptote is $x = -3$ and the horizontal asymptote is $y = 4$.