### Home > APCALC > Chapter 3 > Lesson 3.1.1 > Problem3-18

3-18.

Evaluate each limit. If the limit does not exist due to a vertical asymptote, then add an approach statement stating if $y$ is approaching negative or positive infinity.

1. $\lim\limits_ { x \rightarrow 2 } \large\frac { x ^ { 2 } - x - 2 } { x ^ { 2 } - 3 x + 2 }$

Factor the numerator and denominator. If you can cancel out an $(x−2)$, then the limit exists. If not, then it DNE.

1. $\lim\limits_ { x \rightarrow 5 } \large\frac { \sqrt { x } - \sqrt { 5 } } { x - 5 }$

You could factor the denominator. Or you could multiply the top and bottom by the conjugate of $(\sqrt x − \sqrt5)$.

1. $\lim\limits_ { x \rightarrow 2 ^ { - } } \large\frac { ( 2 x + 1 ) ( x - 5 ) ^ { 6 } } { ( x - 2 ) ^ { 7 } \sqrt { 9 - x } }$

The $(x − 2)$ in the denominator will not cancel out. This means that there is a vertical asymptote at $x = 2$. How can you determine if the asymptote approaches $+\infty$ or $−\infty$ as $x \rightarrow 2$ from the left?

Choose a value that is to left of $x = 2$, and evaluate. Note: You only need to determine if each term is positive or negative. Choose $x \rightarrow 1.9$:

$\lim\limits_{x\rightarrow 1.9}\frac{(2x+1)(x-5)^{6}}{(x-2)^{7}\sqrt{9-x}}=\lim\limits_{x\rightarrow 1.9}\frac{(+)(-)^{6}}{(-)^{7}(\sqrt{+})}$

$=\lim\limits_{x\rightarrow 1.9}\frac{(+)(+)}{(-)(+)}= -\text{ value}$

DNE but $y \rightarrow −\infty$

1. $\lim\limits_ { x \rightarrow \infty } \large\frac { 4 x ^ { 2 } + 2 x + 9 } { - ( 3 x - 6 x ^ { 2 } + 2 ) }$

This is a limit $x \rightarrow \infty$. That means you are looking at end behavior. Is there a horizontal asymptote, or do the $y$-values approach $\infty$ or $−\infty$?

Consider only the terms with the highest power in the numerator and the denominator. Also consider their coefficients, and their signs.