Question

lim sin(x2y)/(x2+y2) (x,y)_(0,0)

Answer 1

解：∵│x^2y/(x^2+y^2)│≤│y│/2
又lim(y->0)y=0
∴lim((x,y)->(0,0))[x^2y/(x^2+y^2)]=lim(y->0)y=0
故 原式=lim((x,y)->(0,0))[(sin(x^2y)/(x^2y))*(x^2y/(x^2+y^2))]
={lim((x,y)->(0,0))[(sin(x^2y)/(x^2y)]}*{lim((x,y)->(0,0))[x^2y/(x^2+y^2)]}
=1*0 (第一个极限应用重要极限lim(z->0)(sinz/z)=1)
=0。

Answer 2

解：∵│x^2y/(x^2+y^2)│≤│y│/2
又lim(y->0)y=0
∴lim((x,y)->(0,0))[x^2y/(x^2+y^2)]=lim(y->0)y=0
故 原式=lim((x,y)->(0,0))[(sin(x^2y)/(x^2y))*(x^2y/(x^2+y^2))]
={lim((x,y)->(0,0))[(sin(x^2y)/(x^2y)]}*{lim((x,y)->(0,0))[x^2y/(x^2+y^2)]}
=1*0 (第一个极限应用重要极限lim(z->0)(sinz/z)=1)
=0。