Difficult Calculus Problem, Can Anyone Figure This Out?

Question

= x^2 + 3, if x < 0
= 0, if x = 0
f(x) = x^2 - 3 if 0 < x < 2
= 1 if x = 2
= x^3 - 7 if 2 < x.

Then lim ... f(x) =
x=>0+


I came up with two answers so far

-3 and 1

Answer 1

I don't know how you got 1, but the answer is -3.

Answer 2

I would guess -3. You are looking at the function at x=0 from the positive side meaning you are finding the limit of x^2-3 as x->0. Put 0 in for x and get the limit as -3.

Answer 3

With a discontinuous function such as this, various limt values are to be expected. Here you have five different branches of the function, so you have several limit points:
x -> 0-, 3.
x -> 0+, -3.
x -> 2-, 1.
x -> 2+, 1
The function is also defined at two isolated points (x = 0 and x = 2); isolated points don't have limits.

Answer 4

-3 cuz it is 0+ , which means nums near 0 but +! and so we should use
f(x) = x^2 - 3 if 0 < x < 2

lim f(0+) = 0-3 = -3