Question

高数极限问题求解~~

确定常数a,b,c使lim(x→1)[(a(x-1)^2+b(x-1)+c-√(x^2+3))/(x-1)^2=0成立。 求详细解答过程 谢谢啦！！

Answer 1

lim(x→1)[(a(x-1)^2+b(x-1)+c-√(x^2+3))/(x-1)^2=0
lim(x→1)[(a(x-1)^2+b(x-1)+c-√(x^2+3))=0
c-2=0,c=2
lim(x→1)[(a(x-1)^2+b(x-1)+2-√(x^2+3))/(x-1)^2=0
罗比达法则
lim(x→1)[(2a(x-1)+b-x/√(x^2+3))/2(x-1)=0
lim(x→1)[(2a(x-1)+b-x/√(x^2+3))=0
b-1/2=0,b=0.5
罗比达法则
lim(x→1)[(2a-[[√(x^2+3)-x^2/√(x^2+3)]/(x^2+3)])/2=0
lim(x→1)[(2a-[[√(x^2+3)-x^2/√(x^2+3)]/(x^2+3)])=0
2a-3/8=0
a=3/16