Question

lim(x,y→2,0)sinxy/[√(xy+1-)1]

求大神解答啊5555。

Answer 1

解：∵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。