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求极限lim(x→0)(1/ln(1+x2)-1/(sinx)2)
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求极限lim(x→0)(1/ln(1+x2)-1/(sinx)2)
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答案和解析
用等价无穷小替换和泰勒公式.
原式=lim(x→0)[sin^2(x)-ln(1+x^2)]/[sin^2(x)ln(1+x^2)]
=lim(x→0)[sin^2(x)-ln(1+x^2)]/(x^2*x^2)
=lim(x→0)[(1-cos(2x))/2-ln(1+x^2)]/x^4
=lim(x→0)[1-cos(2x)-2ln(1+x^2)]/(2x^4)
=lim(x→0)[1-(1-(2x)^2/2!+(2x)^4/4!+o(x^4))-2(x^2-(x^2)^2/2!+o(x^4))]/(2x^4)
=lim(x→0)[1-1+2x^2-2/3x^4+o(x^4)-2x^2+x^4+o(x^4)]/(2x^4)
=lim(x→0)(x^4/3+o(x^4))/(2x^4)
=lim(x→0)1/6+o(1)
=1/6
原式=lim(x→0)[sin^2(x)-ln(1+x^2)]/[sin^2(x)ln(1+x^2)]
=lim(x→0)[sin^2(x)-ln(1+x^2)]/(x^2*x^2)
=lim(x→0)[(1-cos(2x))/2-ln(1+x^2)]/x^4
=lim(x→0)[1-cos(2x)-2ln(1+x^2)]/(2x^4)
=lim(x→0)[1-(1-(2x)^2/2!+(2x)^4/4!+o(x^4))-2(x^2-(x^2)^2/2!+o(x^4))]/(2x^4)
=lim(x→0)[1-1+2x^2-2/3x^4+o(x^4)-2x^2+x^4+o(x^4)]/(2x^4)
=lim(x→0)(x^4/3+o(x^4))/(2x^4)
=lim(x→0)1/6+o(1)
=1/6
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