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求极限.sin(1-x)/lnx(x趁于1);(x^2+2x)/sinx(x趁于0).用等价无穷小代换法求.
人气:279 ℃ 时间:2020-02-05 19:23:59
解答
x->1x-1-->0sin(1-x)~1-x
lnx=ln(1+(x-1))~(x-1)
limsin(1-x)/lnx=-1
x->0sinx~x
lim(x²+2x)/sinx=lim(x²+2x)/x=2
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