大师作文网求下列各极限
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/10 07:46:17
求下列各极限
1、原式=(3-0*1)/(2*0+1^2)=3；
2、∵│x*cos(1/y)│≤│x│,│y*sin(1/x)│≤│y│
又lim(x->0)│x│=0,lim(y->0)│y│=0
∴lm((x,y)->(0,0))[x*cos(1/y)]=0
lm((x,y)->(0,0))[y*sin(1/x)]=0
故 原式=lm((x,y)->(0,0))[x*cos(1/y)]+lm((x,y)->(0,0))[y*sin(1/x)]=0+0=0；
3、原式=lm((x,y)->(0,0)){[4-(xy+4)]/[(xy)(2+√(xy+4))]} (有理化分子)
=lm((x,y)->(0,0)){1/[2+√(xy+4)]}
=1/[2+√(0*0+4)]
=1/4；
4、原式=lm((x,y)->(0,2)){y*[sin(xy)/(xy)]}
=[lm((x,y)->(0,2))(y)]*{lm((x,y)->(0,2))[sin(xy)/(xy)]}
=2*{lm((x,y)->(0,2))[sin(xy)/(xy)]}
=2*[lm(z->0)(sinz/z)] (令z=xy)
=2*1 (应用重要极限lim(t->0)(sint/t)=1)
=2.
2、∵│x*cos(1/y)│≤│x│,│y*sin(1/x)│≤│y│
又lim(x->0)│x│=0,lim(y->0)│y│=0
∴lm((x,y)->(0,0))[x*cos(1/y)]=0
lm((x,y)->(0,0))[y*sin(1/x)]=0
故 原式=lm((x,y)->(0,0))[x*cos(1/y)]+lm((x,y)->(0,0))[y*sin(1/x)]=0+0=0；
3、原式=lm((x,y)->(0,0)){[4-(xy+4)]/[(xy)(2+√(xy+4))]} (有理化分子)
=lm((x,y)->(0,0)){1/[2+√(xy+4)]}
=1/[2+√(0*0+4)]
=1/4；
4、原式=lm((x,y)->(0,2)){y*[sin(xy)/(xy)]}
=[lm((x,y)->(0,2))(y)]*{lm((x,y)->(0,2))[sin(xy)/(xy)]}
=2*{lm((x,y)->(0,2))[sin(xy)/(xy)]}
=2*[lm(z->0)(sinz/z)] (令z=xy)
=2*1 (应用重要极限lim(t->0)(sint/t)=1)
=2.