For f(x) to be continuous at x=0, we require: limx→0−f(x)=f(0)=limx→0+f(x). Given: f(0)=30=0. Right-hand limit:
Consider: limx→0+baxax+b2x2−ax. To simplify, multiply the numerator and the denominator by the conjugate of the numerator: limx→0+bax×(ax+b2x2+ax)(ax+b2x2−ax)×(ax+b2x2+ax). This simplifies to: limx→0+bax×(ax+b2x2+ax)ax+b2x2−ax. Simplifying further: limx→0+bax×(ax+b2x2+ax)b2x2. Canceling terms and simplifying: limx→0+a×2axbx=2ab. For continuity, we equate this limit to the value of f(0): 2ab=3⟹ab=6. Therefore: 6.