Question

高数难题，

求大佬给我解答一下这个问题，是要用重要极限来做吗

Answer 1

用第二个重要极限

Answer 2

利用积化和差、和差化积公式，还有一个重要反常积分：∫(0,+∞)sinx/xdx=π/2 sinxsin(x/3)sin(x/5)=(1/2)*[cos(2x/3)-cos(4x/3)]sin(x/5) =(1/2)*cos(2x/3)sin(x/5)-(1/2)*cos(4x/3)sin(x/5) =(1/4)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)] 原式=(15/4)*∫(0,+∞)[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^3dx =(-15/8)*∫(0,+∞)[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]d(1/x^2) =(-15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2|(0,+∞)+(1/8)*∫(0,+∞)[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x^2dx =lim(x->0) (15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2-(1/8)*∫(0,+∞)[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]d(1/x) =lim(x->0) (15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2-(1/8)*[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x|(0,+∞)-(1/120)*∫(0,+∞)[169sin(13x/15)-49sin(7x/5)-529sin(23x/5)+289sin(17x/15)]/xdx =lim(x->0) {(15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2+(1/8)*[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x}-(π/240)*(169-49-529+289) =π/2+(1/8)*lim(x->0) [15sin(13x/15)-15sin(7x/15)-15sin(23x/15)+15sin(17x/15)+13xcos(13x/15)-7xcos(7x/15)-23xcos(23x/15)+17xcos(17x/15)]/x^2 =π/2+(1/8)*lim(x->0) [13cos(13x/15)-7cos(7x/15)-23cos(23x/15)+17cos(17x/15)-(169/30)xsin(13x/15)+(49/30)xsin(7x/15)+(529/30)xsin(23x/15)-(289/30)xsin(17x/15)]/x =π/2+(1/8)*lim(x->0) [-(169/10)sin(13x/15)+(49/10)sin(7x/5)+(529/10)sin(13x/15)-(289/10)sin(17x/15)] =π/2

追问

不对啊，你做的

Answer 3

把n看为常数

右极限=(C)^(e/x)*e^e=无穷大
左极限=(C)^(e/x)*e^e=0