15Number 15ExplanationIf limx→−31ax+13bx3+27=−135\lim_{x \to -3} \frac{\frac{1}{ax} + \frac{1}{3}}{bx^3 + 27} = -\frac{1}{3^5}limx→−3bx3+27ax1+31=−351, the value of a+ba + ba+b for aaa and bbb positive integers is....−4-4−4−2-2−2000222444Explanationlimx→−31ax+13bx3+27=−135\lim_{x \to -3} \frac{\frac{1}{ax} + \frac{1}{3}}{bx^3 + 27} = -\frac{1}{3^5}x→−3limbx3+27ax1+31=−351 Common denominator in the numerator limx→−33+ax3ax(bx3+27)=−135\lim_{x \to -3} \frac{\frac{3 + ax}{3ax(bx^3 + 27)}} = -\frac{1}{3^5}x→−3lim=3ax(bx3+27)3+ax−351 For x=−3x = -3x=−3 then 3+ax=03 + ax = 03+ax=03−3a=03 - 3a = 03−3a=0a=1a = 1a=1 Substitute x=−3x = -3x=−3 into the denominator bx3+27=0bx^3 + 27 = 0bx3+27=0b(−3)3+27=0b(-3)^3 + 27 = 0b(−3)3+27=0−27b+27=0-27b + 27 = 0−27b+27=0b=1b = 1b=1 Therefore the value of a+b=1+1=2a + b = 1 + 1 = 2a+b=1+1=2.
