limx→1{(x2+ax+b)/[sin(x2-1)]}=3，求a，b的值．

lim_x→1{(x²+ax+b)/[sin(x²-1)]}=3，求a，b的值．
【正确答案】：由于lim_x→1{(x²+ax+b)/[sin(x²-1)]} =lim_x→1[(x²+ax+b)/(x²-1)]=3, 因此一定有lim_x→1{(x²+ax+b)=0，故设x²ax+b=(x-1)(x+c)， 所以3=lim_x→1{[(x-1)(x+c)]/(x²-1)}=lim_x→1[(x+c)/(x+1)]=(1+c)/2, 因此c=5,x²+ax+b=(x-1)(x+5)=x²+4x-5,所以a=4,b=-5.