高中生数学题
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第二题跟第三题方法一样：通分：分子分母同时乘以[(√5x-4)+√x]
分子=(x-1)[(√5x-4)+√x]
分母=4x-4
原=[(√5x-4)+√x] /4=1/2
3、通分母后
分子=(x²+2x+4)-12=(x-2)(x+4)
分母=(x-2)(x²+2x+4)
原式=1/2
4、原==(sinx/cosx-sinx)/x3=sinx/x*[(1-cosx)/x²]=1*(1/2)=1/2
1.∵（x^3-x^2+4x)/(x^2+x)
=(x^2-x+4)/(x+1)
=[(x^2+x)-(2x-4)]/(x+1)
=[x(x+1)-2(x-2)]/(x+1)
=x-2(x-2)/(x+1)
∴原式=lim[x→0][x-2(x-2)/(x+1)]
=lim[x→0]x-lim[x→0]2(x-2)/(x+1)]
=0-2lim[x→0](x-2)/(x+1)]
=2×2
=4.
2.∵(x-1)/[√(5x-4)-√x]
=(x-1)[√(5x-4)+√x]/4(x-1)[分母有理化]
=[√(5x-4)+√x]/4
∴原式=lim[x→1][√(5x-4)+√x]/4
=1/2.
3.∵x^3-8=(x-2)(x^2+2x+4)
∴[1/(x-2)]-[12/(x^3-8)]
=[(x^2+2x+4)-12]/(x^3-8)
=[(x^2+2x-8)/(x-2)(x^2+2x+4)
=(x-2)(x+4)/(x-2)(x^2+2x+4)
=(x+4)/(x^2+2x+4),
∴原式=lim[x→2](x+4)/(x^2+2x+4)
=6/12
=1/2.
4.参考定理:lim[x→0] sinx/x=1,
解法一：
lim[x→0] (tanx-sinx)/x³
=lim[x→0] (sinx/cosx-sinx)/x³
=lim[x→0] (sinx-sinxcosx)/(x³cosx)
=lim[x→0] sinx(1-cosx)/(x³cosx)
=lim[x→0] sin³x(1-cosx)/(x³sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/(sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/[(1-cos²x)cosx]
=lim[x→0] (sinx/x)³·(1-cosx)/[(1+cosx)(1-cosx)cosx]
=lim[x→0] (sinx/x)³·1/[(1+cosx)cosx]
=1·1/(1+1)
=1/2 ；
解法二：
lim[x→0] (tanx-sinx)/x³
=lim[x→0] (sinx/cosx-sinx)/x³
=lim[x→0] (sinx/x)·(1-cosx)/(x²cosx)
=lim[x→0] (sinx/x)·[1-(1-2sin²x/2)]/(x²cosx)
=lim[x→0] (sinx/x)·2sin²(x/2)/(x²cosx)
=lim[x→0] (sinx/x)·2sin²(x/2)/[4(x/2)²cosx]
=lim[x→0] (1/2)·(sinx/x)·[sin(x/2)/(x/2)]²·(1/cosx)
=1/2·1·1²·1
=1/2.
1、把分母搞成 x²+x的关系式
分母=x（x²+x）-2（x²+x）+6x
原=x-2+6/(x+1)=-2
2、分子分母同时乘以[(√5x-4)+√x]
分子=(x-1)[(√5x-4)+√x]
分母=4x-4
原=[(√5x-4)+√x] /4=1/2
3、通分母后
分子=(x²+2x+4)-12=(x-2)(x+4)
分母=(x-2)(x²+2x+4)
原=1/2
4、原=sinx/x*[(1-cosx)/x²]=1*(1/2)²=1/4
第二个得二分之一，第三个得二分之一
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