Limits in Calculus - Please Help Fast

{x^2 + 1, x is greater than or equal to 0}
f(x) = {2x - 3, x is less than 0}
find limit of f(x) as: x approaches 0 fro the left? x approaches 0 from the right? x approaches 0?

Approaches from the left implies x is slightly less than 0.
So, we use f(x) = 2x -3
f(0) = 2(0) - 3 = -3
ANS: -3

Approaches from the right means we use f(x) = x^2 + 1
f(0) = 0^2 + 1 = 1
ANS: 1

Limit as x approaches 0 means we look at the two one-sided limits and see if they agree.
-3 =/= 1 so the limit does not exist.
ANS: DNE

Take x = - h, h --> 0. Then f (x) = -2h - 3 = -3 :::: the left
Take x = h, h --> 0. Then f (x) = 2h - 3 = -3 ::::: the right
Take x = 0. Then f (x) = -3 :::: 0 itself.