09/03/25 14:36:01
>>40
sin(x) ～ x - (1/3!)x^3 + (1/5!) x^5 + O(x^7)
= x { 1 - h(x)} とおくと
(1/x) { {1/sin(x)} - (1/x) }= (1/x^2) { h(x)/(1 -h(x))}
h(x)/x^2 ～ (1/3!) - (1/5!)x^2 + O(x^4)
h(x) → 0 (x→0)
だから
(1/x) { {1/sin(x)} - (1/x) } → 1/3!
09/03/25 14:36:01
>>40
sin(x) ～ x - (1/3!)x^3 + (1/5!) x^5 + O(x^7)
= x { 1 - h(x)} とおくと
(1/x) { {1/sin(x)} - (1/x) }= (1/x^2) { h(x)/(1 -h(x))}
h(x)/x^2 ～ (1/3!) - (1/5!)x^2 + O(x^4)
h(x) → 0 (x→0)
だから
(1/x) { {1/sin(x)} - (1/x) } → 1/3!