I think that you are guilty of sloppy thinking here. We all do that once in a while!And the derivative at x= 0 is that common value.
Actually, what I said was to take the limit of the difference quotient, not of the derivative (though I'd thought of the latter, too). The problem explicitly says to use the definition of the derivative.The limits of \(\displaystyle 6x- 9x^2\) and \(\displaystyle 3x^2+ 6x\) are 0, not 3!
But, as Dr. Peterson said, you were not supposed to take the limits of those but of their derivatives, \(\displaystyle 6- 18x\) and \(\displaystyle 6x+ 6\) which are both 6. And the derivative at x= 0 is that common value, not their sum.