lim(x→-2)(ax+b)/(x+2)=lim(x→-2)[a(x+2)+b-2a]/(x+2)=a+lim(x→-2)(b-2a)/(x+2)由于lim(x→-2)1/(x+2)=∞所以lim(x→-2)(b-2a)/(x+2)为常数的条件是b-2a=0,即b=2a那么lim(x→-2)(ax+b)/(x+2)=a=4,故b=8...