Question

关于求极限

l i m(sin(x-1)/x^2-1)=
x→1

Answer 1

1/2
应用洛必达法则
因为分子分母均为0，所以上下同时取导数
得到：l i m(cos(x-1)/2x)=1/2
x→1
祝学习年愉快～

Answer 2

l i m(sin(x-1)/x^2-1)=
x→1
l i m(sin(x-1)/（（x-1)（x+1））=
x→1
l i m(sin(x-1)/(x-1)*1/(x+1)=
x→1
l i m1/(1+x)=
x→1
1/2.

Answer 3

lim f（x）=f（x0）
x→x0
所以把1代入算出f(1)就行了

Answer 4

sin(x-1)/(x^2-1)=sin(x-1)/[(x-1)(x+1)]

limsin(x-1)/(x^2-1)=limsin(x-1)/(x-1)*lim1/(x+1)
=1*1/2=1/2