  ### Home > APCALC > Chapter 3 > Lesson 3.2.1 > Problem3-51

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.

1. $\lim\limits _ { x \rightarrow - 3 ^ { + } } \frac { x - 3 } { x + 3 }$

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.

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 $−\infty$.

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 $−\infty$.

1. $\lim\limits _ { x \rightarrow 2 } \frac { ( x + 2 ) ^ { 2 } - 2 } { x }$

Evaluate.

1. $\lim\limits _{ x \rightarrow \infty } \frac { 2 + 2 ^ { x } } { 2 - 2 ^ { x } }$

This is a limit $\rightarrow \infty$. Think end behavior. Compare the highest power on the top and bottom. Be sure to consider their coefficients.

$\lim\limits_{x\rightarrow \infty }\frac{2+2^{x}}{2-2^{x}}=\lim\limits_{x\rightarrow \infty }\frac{2^{x}}{-2^{x}}=-1$

1. $\lim\limits _ { x \rightarrow \pi ^ { + } } \pi$ (Careful!)

Think: The function is $y = \pi$. That is a horizontal line. All $y$-values are $\pi$.