Question

求极限lim x→0 y→0 ［1-cos(x²+y²)］/［(x²+y²)e∧x²y²］

Answer 1

e∧x²y² 的极限为1 ，直接 拉倒外面去

［1-cos(x²+y²)］/［(x²+y²)】做变元改为lim(r->0) (1-cosr)/r =0
所以原极限=0

Answer 2

等于limx，y→0 (1/2（x2+y2）)/(x2+y2)e∧x2y2＝limx,y→0 1/(2e∧x2y2)＝1/(2e∧0)＝1/2