Problem #1:
We are approaching x = -1 from the left, and x is less than -1. Thus, use f1(x) = x - 5 instead of f2 or f3.
Just before x reaches -1, the value of f1(x) will be just a bit smaller than -1-5, that is, just a bit smaller than -6. Understanding that x can also equal -1, then the limit of f1(x) as x approaches -1 from the left is -6; it is defined.
Try the next problem. Share your work. I'd be happy to give you feedback on your efforts.
