作业4求极限 a,+c+ lim an ,(a1>0,i=1,2, x→0 解:设y=lma+a2+……+a) In v= lim i Ina+ai In a+.+an Inan) +a+…+ann lim=Eak Inak)=E lim(ak Inag )=lEIna x→0n =-ln(a1a2…an)=ln(ag2…an)n n a,+a+ 故y=lim 作业5若0<λ<1,an>0,且 lim a=a,试证 lim(an+han-1+2-an-2+.+2 ao) …+a 证:lim(an+lan1+12an-2+…+"a)=lim H→) lim 2m+"n+ lin H→)作业 4 求极限 1 1 2 0 lim ,( 0, 1, 2, ) x x x x n i x a a a a i → n ⎧ ⎫ + + + ⎨ ⎬ > = ⎩ ⎭ "" ""n 。 解：设 1 1 2 0 lim x x x x n x a a a y → n ⎧ ⎫ + + + = ⎨ ⎬ ⎩ ⎭ "" ，则 1 2 0 ln ln lim x x xn x a a a n y → x + + + = = " 1 1 1 1 1 2 0 1 ( ln ln ln ) lim 1 x x x x x x n n n x n a a a a a a a a a n → ⋅ + + + + + + " " = 0 0 1 1 1 1 1 lim ( ln ) lim( ln ) ln n n x x k k k k x x k k k a a a a a → → n n n = = ∑ ∑ = = ∑ 1 n k = = 1 1 2 1 2 1 ln( ) ln( )n n n a a a a a a n " " = 故 1 1 2 0 lim x x x x n x a a a y → n ⎧ ⎫ + + + = ⎨ ⎬ ⎩ ⎭ "" = 1 1 2 ( )n n a a "a 。 作业 5 若0 1, n < λ < a > 0，且 lim n n a →∞ = a ，试证： 2 1 2 0 lim ( ) 1 n n n n n a a aλ λ a λ a λ − − →∞ + + + + = − " 。 证： 1 0 1 2 1 2 0 1 1 lim ( ) lim 1 n n n n n n n n n n n a a a a a a λ λ λ λ λ λ − − − − →∞ →∞ + + + a + + + + = " " = 1 1 1 1 1 lim lim 1 1 1 1 n n n n n n n a λ a a λ λ λ λ + + + →∞ →∞ + = = − − − 。 2