用洛必达法则求极限 1,lim(x→0)arctanx-x/sinx^3 2,lim(x→0)lncosax/lncosbx用洛必达法则求极限1,lim(x→0)arctanx-x/sinx^32,lim(x→0)lncosax/lncosbx3,lim(x→0)a^x-x^a/x-a(a>0,a不等于1)3,lim(x→0)(a^x-x^a)/(x-a)(a>0,a不等于1)

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用洛必达法则求极限1,lim(x→0)arctanx-x/sinx^32,lim(x→0)lncosax/lncosbx用洛必达法则求极限1,lim(x→0)arctanx-x/sinx^32,lim

用洛必达法则求极限 1,lim(x→0)arctanx-x/sinx^3 2,lim(x→0)lncosax/lncosbx用洛必达法则求极限1,lim(x→0)arctanx-x/sinx^32,lim(x→0)lncosax/lncosbx3,lim(x→0)a^x-x^a/x-a(a>0,a不等于1)3,lim(x→0)(a^x-x^a)/(x-a)(a>0,a不等于1)
用洛必达法则求极限 1,lim(x→0)arctanx-x/sinx^3 2,lim(x→0)lncosax/lncosbx
用洛必达法则求极限
1,lim(x→0)arctanx-x/sinx^3
2,lim(x→0)lncosax/lncosbx
3,lim(x→0)a^x-x^a/x-a(a>0,a不等于1)
3,lim(x→0)(a^x-x^a)/(x-a)(a>0,a不等于1)
lim(x→0)(e^-2x)/e^x+3x

用洛必达法则求极限 1,lim(x→0)arctanx-x/sinx^3 2,lim(x→0)lncosax/lncosbx用洛必达法则求极限1,lim(x→0)arctanx-x/sinx^32,lim(x→0)lncosax/lncosbx3,lim(x→0)a^x-x^a/x-a(a>0,a不等于1)3,lim(x→0)(a^x-x^a)/(x-a)(a>0,a不等于1)
用洛必达法则求极限
1,lim(x→0)(arctanx-x)/sinx³
x→0lim(arctanx-x)/sinx³=x→0lim[1/(1+x²)-1]/(3x²cosx³)=x→0lim[-x²/(1+x²)(3x²cosx³)]
=x→0lim{-1/[3(1+x²)cosx³]}=-1/3;
如果原题是:x→0lim(arctanx-x)/sin³x,则:
x→0lim(arctanx-x)/sin³x=x→0lim[1/(1+x²)-1]/(3sin²xcosx)=x→0lim{-x²/[3(1+x²)sin²xcosx]}
=x→0lim{-1/[3(1+x²)cosx]}=-1/3;
2,lim(x→0)lncosax/lncosbx
x→0lim[ln(cos(ax)]/[lncos(bx)]=x→0lim[-asin(ax)/cos(ax)]/[-bsin(bx)/cos(bx)]
=x→0lim[atan(ax)/btan(bx)]=x→0lim(a²x)/(b²x)]=a²/b²;
3,lim(x→0)(a^x-x^a)/(x-a)(a>0,a不等于1)
x→0lim(a^x-x^a)/(x-a)=-1/a.

第三题我感觉你题目是不是有问题?第三题不满足可以用洛必达法则的条件啊

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