limx→1=4,则c=_____.
-3 分析:由已知条件,limx→1(x2+2x+c)=0,故(x2+2x+c)=(x-1)(x+k),所以 4=limx→1[(x2+2x+c)/(x-1)]=limx→1[(x-1)(x+k)/(x-1)]=1+k,所以k=3, x2+2x+c=(x-1)(x+3)=(x2+2x-3),即c=-3.
limx→1=4,则c=_____.
-3 分析:由已知条件,limx→1(x2+2x+c)=0,故(x2+2x+c)=(x-1)(x+k),所以 4=limx→1[(x2+2x+c)/(x-1)]=limx→1[(x-1)(x+k)/(x-1)]=1+k,所以k=3, x2+2x+c=(x-1)(x+3)=(x2+2x-3),即c=-3.