lim(x,y→2,0)sinxy/[√(xy+1-)1]
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/09 05:24:12
lim(x,y→2,0)sinxy/[√(xy+1-)1]
∵lim((x,y)->(2,0))[sin(xy)/(xy)]
=lim(t->0)(sint/t) (令t=xy)
=1 (应用重要极限)
lim((x,y)->(2,0))[(xy)/(√(xy+1)-1)]
=lim((x,y)->(2,0))[(xy)(√(xy+1)+1)/(xy+1-1)] (分子分母同乘√(xy+1)+1)
=lim((x,y)->(2,0))[√(xy+1)+1]
=√(2*0+1)+1
=2
∴lim((x,y)->(2,0))[sin(xy)/(√(xy+1)-1)]
=lim((x,y)->(2,0)){[sin(xy)/(xy)]*[(xy)/(√(xy+1)-1)]}
={lim((x,y)->(2,0))[sin(xy)/(xy)]}*{lim((x,y)->(2,0))[(xy)/(√(xy+1)-1)]}
=1*2
=2.
=lim(t->0)(sint/t) (令t=xy)
=1 (应用重要极限)
lim((x,y)->(2,0))[(xy)/(√(xy+1)-1)]
=lim((x,y)->(2,0))[(xy)(√(xy+1)+1)/(xy+1-1)] (分子分母同乘√(xy+1)+1)
=lim((x,y)->(2,0))[√(xy+1)+1]
=√(2*0+1)+1
=2
∴lim((x,y)->(2,0))[sin(xy)/(√(xy+1)-1)]
=lim((x,y)->(2,0)){[sin(xy)/(xy)]*[(xy)/(√(xy+1)-1)]}
={lim((x,y)->(2,0))[sin(xy)/(xy)]}*{lim((x,y)->(2,0))[(xy)/(√(xy+1)-1)]}
=1*2
=2.
数学极限计算lim(x,y)→(0,0) xy/ [√(2-e^xy)-1]= lim(x,y)→(0,0) -xy/(
求下列各极限 lim(x,y)→(0,1) (2-xy)/(x^2+2y)
求极限lim(x,y)→(0,0) [1-cos(xy)]/xy^2.
求极限lim x→0 y→0 2xy/根号下1+xy 然后-1 {不在根号里}
用定义法证明二重极限lim(√(xy+1)-1)/xy=1/2 x,y都趋于0
求极限lim(x→1 y→2) (x²+y²)/xy
求二重极限lim[xy/(1+x^2+y^2)],x→0,y→0.求详细步骤
lim(x,y)→(0,1) (1+xy)^1/x 求大神帮忙,谢谢~
二元函数求极限问题lim[﹙2-e^xy﹚^1/2]-1=lim1/2(1-e^xy)(x,y)→(0,0) (x,y)
xy趋近于0时sinxy/x极限
跪求极限Y=lim (xy+1)/x^4+y^4,当(x,y)→(0,0),
求极限:lim xy分之{[(1+xy)开3次根号]-1},(x,y)→(0,0)