How do you identify the horizontal asymptote for 2/(x+3)?

Feb 23, 2018

$y = 0$

Explanation:

Without using any calculus concepts, such as the limit, here are the general rules for the horizontal asymptotes of a rational function in the form of $P \left(x\right) = \frac{R \left(x\right)}{Q \left(x\right)}$ (which we have here):

1. If the degree of the numerator is less than the degree of the denominator, $y = 0$ is the horizontal asymptote.

2. If the degree of the numerator is greater than the degree of the denominator, we have no horizontal asymptote, but rather a slant asymptote.

3. If the degree of the numerator and the degree of the denominator are equal, the horizontal asymptote is $y = \frac{a}{b}$ where $a$ is the coefficient of the term of highest degree in the numerator and $b$ is the coefficient of the term of highest degree in the denominator.

We have case 1 here (the degree of the numerator is $0$ and the degree of the denominator is $1$); therefore, $y = 0$ is the horizontal asymptote.

With limits:

Take lim_(x->+-∞)f(x):

lim_(x->+∞)(2/(x+3))=2/(∞)=0 (Dividing $2$ by an increasingly large positive number yields $0$ for increasing values of $x$)

lim_(x->+-∞)(2/(x+3))=2/(-∞)=0 (Dividing $2$ by an increasingly large negative number yields $0$ for increasing values of $x$)

Therefore, $y = 0$ is the horizontal asymptote.