The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3. what is the value of the limit as x approaches -3 of [g(x)-g(-3)]over(x+3)?

the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
2y+3=-(2/3)(x-3)
6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = 

2/1 or just 2. sorry if i get this wrong