Question

求极限lim(x+y)sin(1/x²+y²)

Answer 1

1°当(x,y)→(0,0)时
|sin[1/(x²+y²)]|≤1
x+y→0
即
sin[1/(x²+y²)]和 x+y分别为(x,y)→(0,0)时的有界量和无穷小量
故
lim[(x,y)→(0,0)]{(x+y)sin[1/(x²+y²)]}=0

2°当(x,y)→(∞,∞)时
|(x+y)sin[1/(x²+y²)]|
≤ (|x|+|y|)[1/(x²+y²)]
=(|x|+|y|)/(x²+y²)
→0
故
lim[(x,y)→(∞,∞)]{(x+y)sin[1/(x²+y²)]}=0

Answer 2

求极限limx-->0 (x+y)sin(1/x²+y²)
-1<=sin(1/x²+y²)<=1
-limx-->0 (x+y)<=limx-->0 (x+y)sin(1/x²+y²)<=limx-->0 (x+y)
0<=limx-->0 (x+y)sin(1/x²+y²)<=0
∴limx-->0 (x+y)sin(1/x²+y²)=0

Answer 3

如果x、y趋向于0，那么sin(1/x²+y²)是有界函数

lim x+y等于0 一个无穷小乘以一个有界函数极限为0

Answer 4

x,y趋近于何值》