CPM Homework Help : APCALC Problem 3-51

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.

$\lim\limits _ { x \rightarrow - 3 ^ { + } } \frac { x - 3 } { x + 3 }$
Hint (a):
The denominator does not cancel out. That means that there will be a $0$ in the denominator when we evaluate at $x = −3$ , so the limit does not exist.
Hint (a):
Graphically, there will be a vertical asymptote at $x = −3$ . That is why we are asked to find the limit from the right: we want to know if the graph approaches the asymptote towards $+\infty$ or $−\infty$ .
Answer (a):
So evaluate a value that is to the right of $-3$ , and see if you get a positive or negative answer.
$\lim\limits_{x\rightarrow -2.9}\frac{(-)}{(+)}=-\text{value}$
Therefore, from the right, the graph of $y=\frac{x-3}{x+3}$ approaches the vertical asymptote towards $−\infty$ .

$\lim\limits _ { x \rightarrow 2 } \frac { ( x + 2 ) ^ { 2 } - 2 } { x }$
Hint (b):
Evaluate.

$\lim\limits _{ x \rightarrow \infty } \frac { 2 + 2 ^ { x } } { 2 - 2 ^ { x } }$
Hint (c):
This is a limit $\rightarrow \infty$ . Think end behavior. Compare the highest power on the top and bottom. Be sure to consider their coefficients.
Answer (c):
$\lim\limits_{x\rightarrow \infty }\frac{2+2^{x}}{2-2^{x}}=\lim\limits_{x\rightarrow \infty }\frac{2^{x}}{-2^{x}}=-1$

$\lim\limits _ { x \rightarrow \pi ^ { + } } \pi$ (Careful!)
Hint (d):
Think: The function is $y = \pi$ . That is a horizontal line. All $y$ -values are $\pi$ .