例3证明 a>1, lim a= , a<1, 1, 不存在,a=1,a≠1 解设a=r(cos+isin).于是 (1)当|a|<1时,有r<1,则 0<a"=r"(cosn0+sinne)l=r" 因为imrn=0,所以 lim a=o (2)当l>1时,有|<1,则根据()的结论,有
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54
例3证明 a>1, lim a= , a<1, 1 不存在,a=1,a≠1 解设a=r(cos+isin).于是 (1)当|a<1时,先r<1,则 0<a"=r"(cosn0+sinne)l=r" 因为 lim r=0,所以 lim a=o (2)当>1时,先a<,则根据(1)的结论先 所以 lim a
~ 3 y²: limn→∞ α n = 0, |α| > 1, ∞, |α| < 1, 1, α = 1, Ø3, |α| = 1, α 6= 1. ) α = r(cosθ + isinθ). u´ (1) |α| < 1 , k r < 1, K 0 6 |α n | = |r n (cosnθ + sinnθ)| = r n ÏǑ limn→∞ r n = 0, ¤± limn→∞ α n = 0. (2) |α| > 1 , k 1 α < 1, Kâ (1) (Ø, k limn→∞ 1 α n = 0 ¤± limn→∞ α n = ∞. 7/54
(3)当a=1时,有an=1,则
(3) α = 1 , k α n = 1, K limn→∞ α n = 1. (4) |α| = 1, α 6= 1 , k α = cosθ + isinθ, ¤± α n = cosnθ + isinnθ ÏǑ limn→∞ cosnθ Ú limn→∞ sinnθ þØ3, ¤± limn→∞ α n Ø3. 8/54
(3)当a=1时,有an=1,则
(3) α = 1 , k α n = 1, K limn→∞ α n = 1. (4) |α| = 1, α 6= 1 , k α = cosθ + isinθ, ¤± α n = cosnθ + isinnθ ÏǑ limn→∞ cosnθ Ú limn→∞ sinnθ þØ3, ¤± limn→∞ α n Ø3. 8/54
(3)当a=1时,有an=1,则 (4)当a=1,a≠1时,有a=cos+isin0,所以 a"= cosn0+ isinn6
(3) α = 1 , k α n = 1, K limn→∞ α n = 1. (4) |α| = 1, α 6= 1 , k α = cosθ + isinθ, ¤± α n = cosnθ + isinnθ ÏǑ limn→∞ cosnθ Ú limn→∞ sinnθ þØ3, ¤± limn→∞ α n Ø3. 8/54