如题所述
x→1时,ln(2-x)→0,tan(π/2x)→∞,属于0*∞型,把其中一个倒数后放在分母,变成0/0型或者∞/∞型。利用罗必塔法则解题。
原式=lim(x→1)ln(2-x)/[1/tan(π/2x)]
=lim(x→1)ln(2-x)/cot(π/2x)
=lim(x→1)-1/(2-x)/[-π/2csc^2(π/2x)]
=lim(x→1)2sin^2(π/2x)/[(2-x)π]
=sin^2(π/2×1)/[(2-1)π]
=1/π追问
原式=lim(x→1)ln(2-x)/[1/tan(π/2x)]
=lim(x→1)ln(2-x)/cot(π/2x)
=lim(x→1)-1/(2-x)/[-π/2csc^2(π/2x)]
=lim(x→1)2sin^2(π/2x)/[(2-x)π]
=sin^2(π/2×1)/[(2-1)π]
=1/π追问
能手写不,电尹版看不懂
追答
温馨提示:答案为网友推荐,仅供参考