Question

求极限lim(x,y)→(0,0) [1-cos(xy)]/xy^2

求极限lim(x,y)→(0,0) [1-cos(xy)]/xy^2，麻烦写下过程

Answer 1

注：不存在多元函数的罗比达法则。所以，此题不可能有罗比达法则解法。
此题只能转换成一元函数极限求解！
解：原式=lim((x,y)->(0,0))[2sin²(xy/2)/(xy²)] (应用倍角公式)
=lim((x,y)->(0,0))[(x/2)*sin²(xy/2)/(xy/2)²]
={lim((x,y)->(0,0))(x/2)}*{lim((x,y)->(0,0))[sin²(xy/2)/(xy/2)²]}
={lim((x,y)->(0,0))[x/2]}*{lim(z->0)[sinz/z]}² (令z=xy/2)
=(0/2)*1² (应用重要极限lim(z->0)(sinz/z)=1)
=0

Answer 2

0/0型，要用洛必达法则了。
lim((x,y)→(0,0)) (1 - cos(xy))/(xy²)
= lim((x,y)→(0,0)) - [(- ysin(xy)dx) + (- xsin(xy)dy)]/(y²dx + x • 2ydy)
= lim((x,y)→(0,0)) [(xy)sin(xy) + (xy)sin(xy)]/(3xy²)
= (2/3)lim((x,y)→(0,0)) sin(xy)/y
= 2/3 • 0
= 0