求数学极限
limt{[3/(1-x^3)]-[1/(1-x)]}x趋近1参考答案为1limt{[1-cos2x+tan^2(x)]/xsinx}x趋近0参考答案为3我做不出来,求过...
limt{[3/(1-x^3)]-[1/(1-x)]} x趋近1 参考答案为1 limt{[1-cos2x+tan^2(x)]/xsinx} x趋近0 参考答案为3 我做不出来,求过程谢谢各位大佬
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x趋近1时,lim{[3/(1-x^3)]-[1/(1-x)]}
= lim{[3-(1+x+x^2)]/(1-x^3)}
=lim[(2+x)]/(1+x+x^2)]=3/3=1
x趋近0时,(注意因式中,可用等价无穷小x代替sinx)
lim{[1-cos2x+tan^2(x)]/xsinx}
= lim{[1/cos^2(x)-cos2x]/xsinx}
= lim{[1-(1-2sin^2(x))cos^2(x)]/[xsinxcos^2(x)]}
= lim{sin^2(x)[1+2cos^2(x)]/[xsinxcos^2(x)]}
= lim{[(1+2cos^2(x)]x^2/[(x^2)cos^2(x)]}
=3
= lim{[3-(1+x+x^2)]/(1-x^3)}
=lim[(2+x)]/(1+x+x^2)]=3/3=1
x趋近0时,(注意因式中,可用等价无穷小x代替sinx)
lim{[1-cos2x+tan^2(x)]/xsinx}
= lim{[1/cos^2(x)-cos2x]/xsinx}
= lim{[1-(1-2sin^2(x))cos^2(x)]/[xsinxcos^2(x)]}
= lim{sin^2(x)[1+2cos^2(x)]/[xsinxcos^2(x)]}
= lim{[(1+2cos^2(x)]x^2/[(x^2)cos^2(x)]}
=3
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