Question

数学分析，求极限

数学分析，求极限2.5.8.9，能求出几个算几个

Answer 1

(2)
lim(x->0) [e^(2x).sinx -x] /(5x^2+x^3) (0/0)
=lim(x->0) [(2sinx+ cosx).e^(2x) -1] /(10x+3x^2) (0/0)
=lim(x->0) [(4sinx+ 2cosx) + 2cosx -sinx ].e^(2x) /(10+6x)
=lim(x->0) (3sinx+ 4cosx).e^(2x) /(10+6x)
=4
(5)
lim(x->0) ( 1/x^2 - cotx/x )
=lim(x->0) [ 1/x^2 - 1/(x.tanx ) ]
=lim(x->0) ( tanx - x )/(x^2.tanx)
=lim(x->0) ( tanx - x )/x^3 (0/0)
=lim(x->0) [(secx)^2 - 1 ]/(3x^2)
=lim(x->0) (tanx)^2 /(3x^2)
=1/3
(8)
let
y=π-x
lim(x->π) (π-x) tan(x/2)
=lim(y->0) y/tan(y/2)
=2
(9)
lim(x->π/2+) (cosx) ^(π/2-x) =1
lim(x->π/2-) (cosx) ^(π/2-x) 不存在
=>
lim(x->π/2) (cosx) ^(π/2-x) 不存在

Answer 2

代换lim(x÷sinx)在x趋近于0时等于一

Answer 3

2)=lim(2e^2xsinx+e^2xcosx-1)/(10x+3x²)
=lim(4e^2xsinx+4e^2xcosx-e^2xsinx)/(10+6x)
=4/10
5)=lim(1-xcotx)/x²
=lim(-cotx+xcsc²x)/2x
=lim(x-sinxcosx)/2xsin²x
=lim(x-sin2x/2)/2x³
=lim(1-cos2x)/6x²
=lim2sin²x/6x²
=1/3
8)=lim(π-x)/cos(x/2)*limsin(x/2)
=lim-1/-sin(x/2)/2
=2
9)=e^lim(π/2-x)lncosx
=e^limlncosx/(1/(π/2-x))
=e^lim(-sinx/cosx)/(1/(π/2-x)²)
=e^lim-(x-π/2)²/cosx
=e^lim-2(x-π/2)/-sinx
=e^0
=1

追问

第五题，第四步到第五步怎么做的？