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→ TaiBeiGuo :不過難度低一點，引導式的題目 07/06 03:14
推 shaleo :感謝！ 07/08 22:01
※ 引述《shaleo (shaleo)》之銘言：
: 請問
: sin ( x^3 + y^3 )
: lim ----------------------
: x^2 + y^2
: (x,y)->(0,0)
: 應如何解題？
是0喔~~吧!?
Consider
sin ( x^3 + y^3 )
----------------------
x^2 + y^2
sin ( x^3 + y^3 ) x^3 + y^3
= ---------------------- * ----------------------
x^3 + y^3 x^2 + y^2
= A(x,y) * B(x,y)
Since
sinx
lim ─── = 1
t→0 x
so lim A(x,y) = 1
(x,y)→(0,0)
It remains to find lim B(x,y)
(x,y)→(0,0)
Using polar coordinate x=r*cost , y=r*sint
(x,y)→(0,0) <=> r→0 for any t€Real line
so
r^3 * ((cost)^3+(sint)^3)
B(r;t) = B(r*cost,r*sint) = ──────────────
r^2
= r*((cost)^3+(sint)^3)
so for any t€Real line , when r→0
lim B(r;t) = 0 (Since │(cost)^3+(sint)^3│<= 2 , for any t€R)
r→0, any t€R
so 原式 = 1 * 0 = 0
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