1.设f(x)是三次函数,且limx→-1 f(x)/(x+1)=6,limx→-2 f(x)/(x-2)= -3/2,
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1.设f(x)是三次函数,且limx→-1 f(x)/(x+1)=6,limx→-2 f(x)/(x-2)= -3/2,求limx→3 f(x)/(x-3)的值.
2.已知limx→1 (ax^2+bx+1)/(x-1)=3,求limx→∞ ( b^n+a^(n-1) )/( a^n+b^(n-1) )
需要全过程,
2.已知limx→1 (ax^2+bx+1)/(x-1)=3,求limx→∞ ( b^n+a^(n-1) )/( a^n+b^(n-1) )
需要全过程,
1、因为limx→-1 f(x)/(x+1),limx→-2 f(x)/(x-2)存在,所以f(x)必定包含因式(x+1)(x-2),所以设f(x)=A(x+1)(x-2)(x+a).
又因为limx→-1 f(x)/(x+1)=6,limx→2 f(x)/(x-2)= -3/2,所以A(-1-2)(-1+a)=6,A(2+1)(2+a)=-3/2.所以A=1/2,a=-3.所以f(x)=1/2(x+1)(x-2)(x-3).limx→3 f(x)/(x-3)=2
2、因为limx→1 (ax^2+bx+1)/(x-1)存在,所以ax^2+bx+1=(x-1)(ax-1),所以b=-a-1.limx→1 (ax^2+bx+1)/(x-1)=3,所以a-1=3,a=4,b=-5.求limx→∞ ( b^n+a^(n-1) )/( a^n+b^(n-1) )只须将数字代入,并且分子分母同时除以(-5)^n,答案为-5
又因为limx→-1 f(x)/(x+1)=6,limx→2 f(x)/(x-2)= -3/2,所以A(-1-2)(-1+a)=6,A(2+1)(2+a)=-3/2.所以A=1/2,a=-3.所以f(x)=1/2(x+1)(x-2)(x-3).limx→3 f(x)/(x-3)=2
2、因为limx→1 (ax^2+bx+1)/(x-1)存在,所以ax^2+bx+1=(x-1)(ax-1),所以b=-a-1.limx→1 (ax^2+bx+1)/(x-1)=3,所以a-1=3,a=4,b=-5.求limx→∞ ( b^n+a^(n-1) )/( a^n+b^(n-1) )只须将数字代入,并且分子分母同时除以(-5)^n,答案为-5
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