Example 1: Option 1 (Plug it in)

 
$\underset{x\to 5}{\lim }\frac{x^2–6x–16}{x^2+x–72}$ 

Step 1: Try Option 1: Plug it in

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

$\underset{x\to 5}{\lim }\frac{x^2–6x–16}{x^2+x–72}=\frac{{\left(\mathrm{(}5\mathrm{)}\right)}^2–6\left(\mathrm{(}5\mathrm{)}\right)–16}{{\left(\mathrm{(}5\mathrm{)}\right)}^2+\left(\mathrm{(}5\mathrm{)}\right)–72}=\frac{1}{2}$

Final Result:

$\underset{x\to 5}{\lim }\frac{x^2–6x–16}{x^2+x–72}=\frac{1}{2}$

Meaning: The overall limit as $x$approaches $5$ of $\frac{x^2–6x–16}{x^2+x–72}$ is $y=\frac{1}{2}$.