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,
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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,](/uploads/image/z/7247250-18-0.jpg?t=1.%E8%AE%BEf%28x%29%E6%98%AF%E4%B8%89%E6%AC%A1%E5%87%BD%E6%95%B0%2C%E4%B8%94limx%E2%86%92-1+f%28x%29%2F%28x%2B1%29%3D6%2Climx%E2%86%92-2+f%28x%29%2F%28x-2%29%3D+-3%2F2%2C%E6%B1%82limx%E2%86%923+f%28x%29%2F%28x-3%29%E7%9A%84%E5%80%BC.2.%E5%B7%B2%E7%9F%A5limx%E2%86%921+%28ax%5E2%2Bbx%2B1%29%2F%28x-1%29%3D3%2C%E6%B1%82limx%E2%86%92%E2%88%9E+%28+b%5En%2Ba%5E%28n-1%29+%29%2F%28+a%5En%2Bb%5E%28n-1%29+%29%E9%9C%80%E8%A6%81%E5%85%A8%E8%BF%87%E7%A8%8B%2C)
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,
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,
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,
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
1、当x→-1时,(x+1)→0,而limx→-1 f(x)/(x+1)=6,因此必有limx→-1 f(x)=0. 由于三次函数f(x)连续,因此f(-1)=0. 根据因式定理,f(x)有因式(x+1).
同理,由limx→2 f(x)/(x-2)= -3/2知f(x)有因式(x-2).
于是可设三次函数f(x)=(x+1)(x-2)(ax+b). 代入
limx→-1 ...
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1、当x→-1时,(x+1)→0,而limx→-1 f(x)/(x+1)=6,因此必有limx→-1 f(x)=0. 由于三次函数f(x)连续,因此f(-1)=0. 根据因式定理,f(x)有因式(x+1).
同理,由limx→2 f(x)/(x-2)= -3/2知f(x)有因式(x-2).
于是可设三次函数f(x)=(x+1)(x-2)(ax+b). 代入
limx→-1 f(x)/(x+1)=6,limx→2 f(x)/(x-2)= -3/2中,可得方程组
3a-3b=6, 6a+3b=-3/2
解得a=0.5, b=-1.5, f(x)=(x+1)(x+2)(0.5x-1.5).
由此易知
limx→3 f(x)/(x-3)=1.
注:此题条件“limx→-2 f(x)/(x-2)= -3/2”中“limx→-2”应改为“limx→2”,请楼主核对。
2、设f(x)=ax^2+bx+1,仿照上一题,得f(x)有因式(x-1).
可设函数f(x)=(x-1)(cx+d). 于是
ax^2+bx+1=(x-1)(cx+d).
比较两边常数项,有d=-1. 故f(x)=(x-1)(cx-1). 代入题设,得
c-1=3,c=4, f(x)=(x-1)(4x-1)=4x^2-5x+1.
于是a=4,b=-5.
在limn→∞ (b^n+a^(n-1)) / (a^n+b^(n-1))中分子分母同时除以b^(n-1),得
limn→∞ (b+(a/b)^(n-1)) / (a*(a/b)^(n-1)+1)
由于(a/b)^(n-1)=(-4/5)^(n-1)→0, 因此上述极限式中,分子趋于-5, 分母趋于1,极限为-5.
注:此题所求“limx→∞ ( b^n+a^(n-1) )/( a^n+b^(n-1) )
”中“limx→∞”应改为“limn→∞”,请楼主核对。
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