![1、lim x->0 [(1+x)^1x-(1+2x)^12x]sinx 2、lim x->0 (1x^2-1(arctanx)^2),第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw== "1、lim x->0 [(1+x)^1x-(1+2x)^12x]sinx 2、lim x->0 (1x^2-1(arctanx)^2),第1張")
1、lim x->0 [(1 x)^1/x-(1 2x)^1/2x]/sinx 2、lim x->0 (1/x^2-1/(arctanx)^2)
1、e/2 2、 -2/3網友廻答：
- 樓上犯的是典型錯誤.
1、根據洛必達法則
lim( x->0) [(1 x)^1/x-(1 2x)^1/2x]/sinx =lim (x->0){ [(1 x)^1/x]g(x)-[(1 2x)^1/2x]h(x)}/cosx
這裡g(x)=-(1/x^2)ln(1 x) 1/[x(1 x)]
h(x)=(1/2){[-(1/x^2)ln(1 2x)] 2/[x(1 2x)]}
lim( x->0)g(x)= lim( x->0){[-(1/x^2)ln(1 x)] 1/[x(1 x)]}=lim( x->0)[-ln(1 x)/2x]=-1/2
lim( x->0)h(x)= lim( x->0)(1/2){[-(1/x^2)ln(1 2x)] 2/[x(1 2x)]}
=lim( x->0)(1/2)[-2ln(1 2x)/2x]=-1
∴原式=e(-1/2)-e(-1)=e/2
2、
lim( x->0) (1/x^2-1/(arctanx)^2)
=lim(x->0)(arctanx²-x²)/(x²arctanx²)=lim(x->0)(arctanx²-x²)/(x^4)
=lim(x->0)[(2arctanx)/(1 x^2)-2x]/4x^3=lim(x->0)[(arctanx-x-x^3]/2x^3
=lim(x->0){[1/(1 x^2)]-1-3x^2}/6x^2=lim(x->0)(-4x^2-3x^4)/6x^2=-2/3
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![1、lim x->0 [(1+x)^1x-(1+2x)^12x]sinx 2、lim x->0 (1x^2-1(arctanx)^2),第2張](/statics/default/images/niming.jpg "1、lim x->0 [(1+x)^1x-(1+2x)^12x]sinx 2、lim x->0 (1x^2-1(arctanx)^2),第2張")
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