If a function is odd then f(-x) = -f(x), if a the function is even, then f(-x) = f(x). In this case we have x = 0, meaning (and an odd function), which means f(-0) = -f(0) --> f(0) = -f(0). There is only one way for this to be true: f(0) = 0, such that -0 = +0. Notice that this is NOT true for an even function: f(-0) = f(0) --> f(0) = f(0) --> f(0) could be ANYTHING (as an example: f(x) = x^2 + 3 such that f(0) = 3 and f(-0) = 3 --> f(0) = f(-0) = 3, but it's NOT odd because f(-0) != -f(0) --> 3 != -3).
Therefore, f(0) = 0, therefore 2 - 3*f(0) = 2 - 3*0 = 2
Edit: there is a part of the question which is important: that the limit exists when x approaches zero. It's NOT true that an odd function MUST pass through the origin. It could very well be undefined at the origin. Furthermore, it's NOT necessary that it's limit exist at zero either. Here is an example of an odd function whose limit does not exist at x = 0:
f(x) = {ceiling(|x|), x > 0
f(x) = {-ceiling(|x|), x < 0
f(x) = {0, x = 0
This function is defined for ALL real values but has jump discontinuities at all integer values of x. Furthermore, while f(0) is defined (to be zero), the limit as x approaches zero does not exist (as it will be -1 from the left and +1 from the right).
Second Edit: Perhaps an even more intuitive example is a hyperbola:
f(x) = 1/x<-- this too is an odd function (but whose limit as x approaches zero doesn't exist).