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 $+∞$ or $−∞$.

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 $−∞$

Evaluate.

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

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

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