高数极限10道题求解和过程

Question

如题所述

Answer 1

(1)
lim(x->-2) (x-2)/(x^2-1)
=(-2-2)/(4-1)
=-4/3
(2)
lim(x->π/2) ln(1+cosx)/sinx
=ln(1+0)/1
=0
(3)
lim(x->+∞) (x-1)(x-2)(x-3)/( 1- 4x)^3
分子分母同时除以x^3
=lim(x->+∞) (1-1/x)(1-2/x)(1-3/x)/( 1/x- 4)^3
=1/(-4)^3
=-1/64
(4)
let
y=1/x
lim(x->+∞) x. tan(1/x)
=lim(y->0) tany /y
=1
(5)
lim(x->0) (1-cosx)/(xsinx)
=lim(x->0) (1/2)x^2)/x^2
=1/2
(6)
应该是
lim(x->1) [ 1/(1-x) -3/(1-x^3) ]
=lim(x->1) (1+x+x^2-3)/[(1-x)(1+x+x^2)]
=lim(x->1) (x^2+x-2)/[(1-x)(1+x+x^2)]
=lim(x->1) (x-1)(x+2)/[(1-x)(1+x+x^2)]
=lim(x->1) -(x+2)/(1+x+x^2)
=-1
题目
lim(x->1) [ 1/(1-x) -1/(1-x^3) ]
=lim(x->1) (1+x+x^2-1)/[(1-x)(1+x+x^2)]
=lim(x->1) (x^2+x)/[(1-x)(1+x+x^2)]
->∞
(7)
lim(x->0) sinx. cos(1/x)
|cos(1/x)|<=1
lim(x->0) sinx =0
=>
lim(x->0) sinx. cos(1/x) =0
(8)
x->0
sinx~ x
tanx ~x
lim(x->0) (1+sinx)^tanx
=lim(x->0) (1+x)^x
= (1+0)^0
=1
(9)
L =lim(x->+∞) [x^2/(x^2-1)]^x
lnL
=lim(x->+∞) xln[x^2/(x^2-1)]
=lim(x->+∞) ln[x^2/(x^2-1)] /(1/x) (0/0 分子分母分别求导）
=lim(x->+∞) [2/x - 2x/(x^2-1)] /(-1/x^2)
=lim(x->+∞) 2x^2 /[ x(x^2-1) ]
=0
=> L =1
lim(x->+∞) [x^2/(x^2-1)]^x =1
(10)

lim(x->0+) (lnx)^x 不存在

Answer 2