Example 2: Factor and Cancel

 
$\underset{\mathrm{t}\to –3}{\lim }\frac{\left(\mathrm{t}+3\right)}{\left({\mathrm{t}}^2–9\right)}$ 

Step 1: Try Option 1: Plug it in

Always try plugging in the $x–\mathit{value}$ you are heading towards.

In this example we get division by zero, which means Option 1 has failed.

$\underset{t\to –3}{\lim }\frac{\left(t+3\right)}{\left(t^2–9\right)}=\frac{\left((–3)+3\right)}{\left({(–3)}^2–9\right)}=\frac{0}{0}$

Step 2: Factor

The factoring really is the hard part. There is no quick and easy process to factoring. The more you practice factoring the better you will become.

IMPORTANT: This problem is using a very common factoring situation known as the difference of two squares . The factoring of this type of equation always works out the same. Make sure you can identify it, and use it. It will show up ALL the time.

$\underset{t\to –3}{\lim }\frac{\left(t+3\right)}{\left(t^2–9\right)}=\underset{t\to –3}{\lim }\frac{\left(t+3\right)}{\left(\mathrm{(}t+3\mathrm{)}\right)\left(\mathrm{(}t–3\mathrm{)}\right)}$

Step 3: Cancel

Cancel out any common factors between the top and bottom of the fraction.

$\underset{\mathrm{t}\to –3}{\lim }\frac{\overline{)\left(\mathrm{(}\mathrm{t}+3\mathrm{)}\right)}}{\overline{)\left(\mathrm{(}\mathrm{t}+3\mathrm{)}\right)}\left(\mathrm{(}\mathrm{t}–3\mathrm{)}\right)}=\underset{t\to –3}{\lim }\frac{1}{\left(\mathrm{(}t–3\mathrm{)}\right)}$

Step 4: Back to Option 1: Plug it in

$\underset{t\to –3}{\lim }\frac{1}{\left(\mathrm{(}t–3\mathrm{)}\right)}=\frac{1}{\left(\mathrm{(}–3\mathrm{)}\right)–3}=–\frac{1}{6}$

Final Result:

$\underset{\mathrm{t}\to –3}{\lim }\frac{\left(\mathrm{t}+3\right)}{\left({\mathrm{t}}^2–9\right)}=–\frac{1}{6}$

Meaning:

1)      The overall limit as $t$approaches $–3$ of $\frac{\left(\mathrm{t}+3\right)}{\left({\mathrm{t}}^2–9\right)}$ is $y=–\frac{1}{6}$.

2)       There is a removeable discontinuity (hole in the graph) at $t=–3$.

Where the canceled factor, $\left(\mathrm{(}t+3\mathrm{)}\right)$, equals zero.