洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/18 21:48:49
洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)
xXNG*UnG` O". JihPZQp@> ٽ?vBJ-}{7;LL= OtNNZG=zٛ5YIKvtN7_yG@z>ukdA$zSzR՚o%gj[9Gt|nGJ y%P1ז")lߕ`I^w+gu 8'8}4o 2=" wfr?~|hIJH-8=.[j -AR18+6'ψ̑lr6!%%dÑȾ*&\̤99 o-MS(.Vf#*6_||޻h;6Lorӗ+_=;KZfUR  pLG[2f9Jy%( h )F d(;$GAESy0k+$ xyr <'|h dEΨ4Fprs&ʚj-]s!T YSif 35YUh9 @9 L&7/g͵vt]xh9SQuuz>獻!CЭƒXSOzäϨ31JX/\1l|_LqyZ}`ԡ%JѹS\# _xAx6ۇ7@i:?^geK;.|ÝAs L;o<4q(,/j%";(A(L"Aih݆y5ȴ o#VzҀP #X\*e{RP^<)hJ&'W)d\D%.o`Pgõ7x&h@\Cэq2S.Í:5Ջ^jJ+HƺiQ<wK,Xb/ՠ x dt(aD PYÔ3)Fլ*V|ru=D5 *`FwQGtfW*;!WFSZ07P{\fY2a\V܆֕X~n_d`]Z" $0Ł߰!j;ɑ2(j.q l)xc >Nkyz

洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)
洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,
limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)^(1/x)]^(1/x) x趋近0

洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)
1.lnx*ln(1+x)=ln(1+x)/(1/lnx)
=[1/(1+x)][-1/(lnx)^2*1/x]
=x(lnx)^2/(1+x)
=(lnx)^2/(1+1/x)
=[2lnx/x]/(-1/x^2)
=-2lnx/(1/x)
=(-2/x)/(-1/x^2)
=2x=0
2.(x^3+x^+x+1)^1/3-x
=[(x^3+x^+x+1)^1/3-x]
*[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
=[x^3+x^2+x+1-x^3]/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
=(x^2+x+1)/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
上下同除x^2
=(1+1/x+1/x^2)/[(1+1/x+1/x^2+1/x^3)^(2/3)+(1+1/x+1/x^2+1/x^3)^(1/3)+1]
=(1+0+0)/(1+1+1)=1/3
3.(e^x-e^sinx)/(x-sinx)
=(e^x-e^sinx *cosx)/(1-cosx)
=(e^x-e^sinx *(cosx)^2+e^sinx*cosx*sinx)/sinx
=(e^x-e^sinx *(cosx)^3+2e^sinx *cosx sinx+e^sinx*(cosx)^2*sinx+e^sinx*(cos 2x))/cosx
=(1-1+0+0+1)/1=1
4.令y=(sinx/x)^(1/x^2)
lny=(ln sinx -ln x)/x^2
=(cosx/sinx -1/x)/2x
=(-csc^2 x +1/x^2)/2
=(1/x^2-1/sinx^2)/2
=[(sinx)^2-x^2]/(2x^2 sinx^2)
=[2sinx cosx -2x]/[4x sinx^2+4x^2sinx cosx]
=[sin2x-2x]/[2x(1-cos 2x)+2x^2 sin 2x]
=[2cos 2x -2]/[2(1-cos 2x)+4x sin 2x +4x sin2x +4x^2cos 2x]
=[cos2x-1]/[1-cos2x+4xsin2x+2x^2cos2x]
=[-2sin2x]/[2sin2x+4sin2x+8xcos2x+4xcos2x-4x^2sin2x]
=-4cos2x/[4cos2x+8cos2x+8cos2x-16xsin2x+4cos2x-8xsin2x-8xsin2x-8x^2cos2x]
=-4/[4+8+8+0+4-0-0-0]
=-4/24=-1/6
y->exp(-1/6)
5.令y=[1/e· (1/+x)^(1/x)]^(1/x)
lny=ln[1/e· (1/+x)^(1/x)]/x
=[(ln(1+x))/x-1]/x
=[ln(1+x)-x]/x^2
=[1/(1+x)-1]/2x
=[-1/(1+x)^2]/2
=-1/2
y->e^(-1/2)

说明:这8道题有很多种简便的解法。但如果要求用罗比达法则求解,那就是如下解法。
(1)lim(x->0+)(sinx*lnx)=?
原式=lim(x->0+)(x*lnx) (应用等价代换sinx~x)
=lim(x->0+)[lnx/(1/x)]
=l...

全部展开

说明:这8道题有很多种简便的解法。但如果要求用罗比达法则求解,那就是如下解法。
(1)lim(x->0+)(sinx*lnx)=?
原式=lim(x->0+)(x*lnx) (应用等价代换sinx~x)
=lim(x->0+)[lnx/(1/x)]
=lim(x->0+)[(1/x)/(-1/x²)] (∞/∞型极限,应用罗比达法则)
=lim(x->0+)(-x)
=0;
(2)lim(x->∞)[((2/π)arctanx)^x]=?
原式=e^{lim(x->∞)[(ln(arctanx)-ln(2/π))/(1/x)]}
=e^{lim(x->∞)[((1/arctanx)(1/(1+x²)))/(-1/x²)]} (0/0型极限,应用罗比达法则)
=e^{lim(x->∞)[(-1)/((1+1/x²)arctanx)]}
=e^[(-1)/((1+0)(π/2))]
=e^(-2/π);
(3)lim(x->0+)[(ln(1/x))^x]=?
原式=e^{lim(x->0+)[ln(1/x)/(1/x)]}
=e^{lim(x->0+)[(x(-1/x²))/(-1/x²)]} (∞/∞型极限,应用罗比达法则)
=e^{lim(x->0+)(x)}
=e^(0)
=1;
(4)lim(x->0+)[lnx*ln(1+x)]=?
原式=lim(x->0+)(x*lnx) (应用等价代换ln(1+x)]~x)
=lim(x->0+)[lnx/(1/x)]
=lim(x->0+)[(1/x)/(-1/x²)] (∞/∞型极限,应用罗比达法则)
=lim(x->0+)(-x)
=0;
(5)lim(x->∞)[(x³+x²+x+1)^(1/3)-x]=?
原式=lim(x->∞)[x((1+1/x+1/x²+1/x³)^(1/3)-1)]
=lim(x->∞)[((1+1/x+1/x²+1/x³)^(1/3)-1)/(1/x)]
=lim(x->∞)[(1/3)(1+2/x+2/x²)/(1+1/x+1/x²+1/x³)^(2/3)] (0/0型极限,应用罗比达法则)
=(1/3)(1+0+0)/(1+0+0+0)^(2/3)
=1/3;
(6)lim(x->0)[(e^x-e^(sinx))/(x-sinx)]=?
原式=lim(x->0)[e^(sinx)(e^(x-sinx)-1)/(x-sinx)]
=lim(x->0)[e^(sinx)]*lim(x->0)[(e^(x-sinx)-1)/(x-sinx)]
=1*lim(x->0)[(e^(x-sinx)-1)/(x-sinx)]
=lim(x->0)[(e^(x-sinx)] (0/0型极限,应用罗比达法则)
=e^(0-0)
=1;
(7)lim(x->0)[(sinx/x)^(1/x²)]=?
原式=e^{lim(x->0)[(ln(sinx)-lnx)/x²]}
=e^{lim(x->0)[(cosx/sinx-1/x)/(2x)]} (0/0型极限,应用罗比达法则)
=e^{lim(x->0)[(xcosx-sinx)/(2x²sinx)]}
=e^{lim(x->0)[(-sinx)/(2xcosx+4sinx)]} (0/0型极限,应用罗比达法则)
=e^{lim(x->0)[-(sinx/x)/(2cosx+4(sinx/x))]}
=e^(-1/(2+4)) (应用重要极限lim(x->0)(sinx/x)=1)
=e^(-1/6);
(8)lim(x->0)[((1/e)(1+x)^(1/x))^(1/x)]=?
原式=e^{lim(x->0)[(ln(1+x)/x-1)/x]}
=e^{lim(x->0)[(ln(1+x)-x)/x²]}
=e^{lim(x->0)[(-1)/(2(1+x))]} (0/0型极限,应用罗比达法则)
=e^(-1/(2(1+0)))
=e^(-1/2)。

收起