Question

详细过程，最好拍下来。高数求极限

Answer 1

Answer 2

1、原式=lim(x->0) (1-x^2*cot^2x)/x^2……通分
=lim(x->0) (tan^2x-x^2)/(tan^2x*x^2)
=lim(x->0) (tan^2x-x^2)/x^4……分母用等价无穷小代换
=lim(x->0) (tanx*sec^2x-x)/2x^3……洛必达法则
=lim(x->0) (sec^4x+2tan^2x*sec^2x-1)/6x^2
=lim(x->0) (3sec^4x-2sec^2x-1)/6x^2……三角恒等式变换
=lim(x->0) (3sec^2x+1)(sec^2x-1)/6x^2
=(2/3)*lim(x->0) tan^2x/x^2
=(2/3)*lim(x->0) x^2/x^2……分子用等价无穷小代换
=2/3
2、原式=e^lim(x->0+) [ln(arcsinx/x)]/(1-cosx)
=e^lim(x->0+) {ln[(x+x^3/6+o(x^4))/x]}/(x^2/2)……分母用等价无穷小代换，分子用arcsinx的泰勒展开
=e^lim(x->0+) 2[ln(1+x^2/6+o(x^3)]/x^2
=e^lim(x->0+) 2[x^2/6+o(x^3)]/x^2……分子用等价无穷小代换
=e^lim(x->0+) 2[1/6+o(x^3)/x^2]
=e^(1/3)

追问

十分清晰完整，非常感谢。

追答

客气~