Question

求解答，要过程。谢谢！

Answer 1

1.

(1)

lim(x->0) (x^3+2x+5)/(x+1)

=(0+0+5)/(0+1)

=5

(2)

lim(x->2) (x^2+x-6)/(x^2-4)

=lim(x->2) (x-2)(x+3)/[(x-2)(x+2)]

=lim(x->2) (x+3)/(x+2)

=(2+3)/(2+2)

=5/4

(3)

lim(x->0) [√(x^2+1)-1 ]/x

分子分母同时乘以 [√(x^2+1)+1 ]

=lim(x->0) x^2/{x.[√(x^2+1)+1 ] }

=lim(x->0) x/[√(x^2+1)+1 ] 

=0

(4)

lim(x->0) (4x^3-2x^2+x)/(3x^2+2x)

=lim(x->0) (4x^2-2x+1)/(3x+2)

=(0-0+1)/(0+2)

=1/2

(5)

lim(x->∞) (1+1/x)(2-1/x^2)

=(1+0)(2-0)

=2

(6)

lim(x->∞) cosx/[e^x +e^(-x)]

分子分母同时除e^x

=lim(x->∞) [cosx/e^x]/[1 +1/e^(2x)]

=0/(1+0)

=0

2.

(1)

lim(x->0) sinkx/x

=lim(x->0) kx/x

=k

(2)

lim(x->0) sin5x/(3x)

=lim(x->0) 5x/(3x)

=5/3

Answer 2

1. (1)lim[x-->0](x^2+2x+5)/(x+1)=5 --------------------------------------- 代入法
   

   (2)lim[x-->2](x^2+x-6)/(x^2-4)=lim[x-->2](x+3)/(x+2)=5/4---------同除（x-2）

   (3)lim[x-->0](√(x^2+1)-1)/x   ---------------------------------------同乘(√(x^2+1)+1)
   

   =lim[x-->0][(√(x^2+1)-1)(√(x^2+1)+1)]/[x(√(x^2+1)+1)]-------有理化

   =lim[x-->0]x/(√(x^2+1)+1)=0

   (4)lim[x-->0](4x^3-2x^2+x)/(3x^2+2x)   --------------------------分子分母同除x
   

   =lim[x-->0](4x^2-2x+1)/(3x+2)=1/2

   (5)lim[x-->∞0](1+1/x)(2-1/x^2)=2   ---------------------------利用无穷大与无穷小关系
   

   (6)∵ |cosx|<=1,lim[x-->+∞]1/(e^x+e^(-x))=0 

   ∴ lim[x-->+∞]cosx/(e^x+e^(-x))=0     --------------------无穷小的性质
   

2. (1)lim[x-->0]sinkx/x=k                            -----------------------等价无穷小代换
   

   (2)lim[x-->0]sin5x/3x=5/3      -----------------------等价无穷小代换
   

Answer 3

解答如下

Answer 4