设函数f(x,y)=xsin1/y+ysin1/x,则极限z=e^xsiny
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lim x→0y→0(xsin1/y+ysin1/x)
= lim x→0y→0(xsin1/y + xsin1/x - xsin1/x + ysin1/x)
= lim x→0y→0[x(sin1/y + sin1/x)] + lim x→0y→0 [(y- x)sin1/x]
= 0 + 0
= 0
（因为 -2 ≤ sin1/y + sin1/x ≤ 2
-1 ≤ sin1/x ≤ 1
所以 lim x→0y→0[x(sin1/y + sin1/x)] = 0
lim x→0y→0 [(y- x)sin1/x] = 0 ）
= lim x→0y→0(xsin1/y + xsin1/x - xsin1/x + ysin1/x)
= lim x→0y→0[x(sin1/y + sin1/x)] + lim x→0y→0 [(y- x)sin1/x]
= 0 + 0
= 0
（因为 -2 ≤ sin1/y + sin1/x ≤ 2
-1 ≤ sin1/x ≤ 1
所以 lim x→0y→0[x(sin1/y + sin1/x)] = 0
lim x→0y→0 [(y- x)sin1/x] = 0 ）
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