If the -8 is under the square root, then...
![\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%20-3%7D%20%5Csqrt%7Bx%5E2-8%7D%5C%5C%5C%5CL%20%3D%20%5Csqrt%7B%28-3%29%5E2-8%7D%5C%5C%5C%5CL%20%3D%20%5Csqrt%7B9-8%7D%5C%5C%5C%5CL%20%3D%20%5Csqrt%7B1%7D%5C%5C%5C%5CL%20%3D%201%5C%5C%5C%5C)
OR
If the -8 is not under the square root, then...
![\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%20-3%7D%20%5Csqrt%7Bx%5E2%7D-8%5C%5C%5C%5CL%20%3D%20%5Csqrt%7B%28-3%29%5E2%7D-8%5C%5C%5C%5CL%20%3D%20%5Csqrt%7B9%7D-8%5C%5C%5C%5CL%20%3D%203-8%5C%5C%5C%5CL%20%3D%20-5)
Either way, we replace x with -3 and simplify.
For more information, refer to the direct substitution rule for limits.
A I guess (finger crossed )
Answer:
yes very much so
Step-by-step explanation:
yes I don't know why but I am really and truly
B and D is correct
PS QT
GOOD LUCK