Example 1: Factor and Cancel

$\lim_{x \to 8} \frac{x^{2} - 6 x - 16}{x^{2} + x - 72}$

Step 1: Try Option 1: Plug it in

Always try plugging in the $x - value$ you are heading towards.

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

$\lim_{x \to 8} \frac{x^{2} - 6 x - 16}{x^{2} + x - 72} = \frac{{((8))}^{2} - 6 ((8)) - 16}{{((8))}^{2} + ((8)) - 72} = \frac{0}{0}$

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.

In this example we are able to factor both the top and the bottom of the fraction.

$\lim_{x \to 8} \frac{x^{2} - 6 x - 16}{x^{2} + x - 72} = \lim_{x \to 8} \frac{((x - 8)) ((x + 2))}{((x - 8)) ((x + 9))}$

Step 3: Cancel

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

$\lim_{x \to 8} \frac{((x - 8)) ((x + 2))}{((x - 8)) ((x + 9))} = \lim_{x \to 8} \frac{((x + 2))}{((x + 9))}$

Step 4: Back to Option 1: Plug it in

$\lim_{x \to 8} \frac{((x + 2))}{((x + 9))} = \frac{((8)) + 2}{((8)) + 9} = \frac{10}{17}$

Final Result:

$\lim_{x \to 8} \frac{x^{2} - 6 x - 16}{x^{2} + x - 72} = \frac{10}{17}$

Meaning:

1) The overall limit as $x$ approaches $8$ of $\frac{x^{2} - 6 x - 16}{x^{2} + x - 72}$ is $y = \frac{10}{17}$ .

2) There is a removeable discontinuity (hole in the graph) at $x = 8$ .

Where the canceled factor, $((x - 8))$ , equals zero.