Question

lim(x→1)[a(x-1)^2+b(x-1)+c-√(x^2+3)]/(x-1)^2=0 确定 a.b.c的值 不用

Answer 1

要使得当x→1时,[a(x-1)^2+b(x-1)+c-√(x^2+3)]/(x-1)^2的极限存在,则当x→1时,a(x-1)^2+b(x-1)+c-√(x^2+3)]→0,所以c＝2,代入得[a(x-1)^2+b(x-1)+c-√(x^2+3)]/(x-1)^2＝[a(x-1)^2+b(x-1)+2-√(x^2+3)]/(x-1)^2＝[a(x-1)+b－(1+x)/(2+√(x^2+3))]/(x-1),要使得极限存在,则x→1时,a(x-1)+b－(1+x)/(2+√(x^2+3))→0,则b＝1/2,代入并化简得：a(x-1)+b－(1+x)/(2+√(x^2+3))＝a(x-1)+1/2－(1+x)/(2+√(x^2+3))＝[a－3(1+x)/2(2+√(x^2+3))(√(x^2+3)+2x)](x-1),故有当x→1时,a－3(1+x)/2(2+√(x^2+3))(√(x^2+3)+2x)→0,所以a＝1/6.