If the -8 is under the square root, then...
[tex]\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\[/tex]
OR
If the -8 is not under the square root, then...
[tex]\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5[/tex]
Either way, we replace x with -3 and simplify.
For more information, refer to the direct substitution rule for limits.
