Question

求lim(x趋向于1)(x^2-5x+2)/(x^3-4x+1)的极限，要详细过程！谢谢！

要先把式子给解开

Answer 1

1、
x→+∞
lim (4x^2-3x+1) / (2x^2-6x+5)
=lim (4x^2-3x+1)/x^2 / (2x^2-6x+5)/x^2
=lim (4-(3/x)+(1/x^2)) / (2-(6/x)+(5/x^2))
=(4-0+0) / (2-0+0)
=4/2
=2
2、
x→+∞
lim [(2x-1)/(2x+1)]^(x+1)
=lim [(2x+1-2)/(2x+1)]^(x+1)
=lim [1-2/(2x+1)]^(x+1)
=lim [1-2/(2x+1)]^[2(x+1)/2]
=lim [1-2/(2x+1)]^[(2x+1+1)/2]
=lim [1-2/(2x+1)]^[(2x+1)/2] * lim [1-2/(2x+1)]^[1/2]
=lim [1-2/(2x+1)]^[(2x+1)/2] * 1
=lim [1-2/(2x+1)]^[(2x+1)/-2 * -1]
=[lim [1-2/(2x+1)]^[(2x+1)/-2]]^(-1)
根据重要的极限：lim(x→∞) (1+1/x)^x=e

Answer 2

如图

Answer 3

lim(x->1) (x^2-5x+2)/(x^3-4x+1) = -2/-2 = 1