Question

求极限lim(x+y)sin(1/x²+y²)

Answer 1

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

2°当(x,y)→(∞,∞)时
|(x+y)sin[1/(x²消薯+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闹猜祥+y²)
-1<=sin(1/x²+y²)<=1
-limx-->0 (x+y)<液搏=limx-->0 (x+y)sin(1/x²+y²)<=limx-->0 (x+y)
0<=limx-->0 (x+y)sin(1/x²兆扮+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)是有界函者没知数

lim x+y等于0 一个无穷小乘以一察脊个有界首消函数极限为0

Answer 4

x,y趋近于何值》