Question

求极限: x趋于1,(x^2-3x+2)/(x^3-4x+3)

详解,谢谢

Answer 1

由洛必达法则有
Lim[(x^2 - 3 x + 2)/(x^3 - 4 x + 3), x -> 1]
=Lim[(x^2 - 3 x + 2)`/(x^3 - 4 x + 3)`, x -> 1]
=Lim[(2x-3)/(3x^2-4), x -> 1]
=1

洛必达法则:
由洛必达法则有
Lim[g(x)/f(x), x -> x0]
=Lim[g`(x)/f`(x), x -> x0]
应用于0/0型和∞/∞型的极限.

另解:
Lim[(x^2-3x+2)/(x^3-4x+3),x->1]
=Lim{[(x-2)(x-1)]/[(x-1)(x^2+x-3)],x->1}
=Lim[(x-2)/(x^2+x-3),x->1]
=1

Answer 2

(x^2-3x+2)/(x^3-4x+3)=[(x-2)(x-1)]/[(x-1)(x^2+x-3)]
=(x-2)/(x^2+x-3)
x趋于1
所以把1代入,极限为:1

Answer 3

(x^2-3x+2)/(x^3-4x+3)
=(x-1)(x-2)/(x-1)(x^2+x-1)
=(x-2)/(x^2+x-1)

x趋于1时
上式趋于-1

Answer 4

零。=（X-1）（X-2）/X（X+2）（X-2）+3
=（X-1）/X（X+2）+3/（X-2），当X=1，枝为零

Answer 5

相消再带入比较简单...