六分之一x的三次方是怎么来的?
追答泰勒展式
f(x) = f(0) +[f'(0)/1!]x + [f''(0)/2!]x^2 +.... + [f^(n)(0)/n!] x^n +....
f(x) = sinx => f(0) =0
f'(x) = cosx => f'(0)/1! =1
f''(x) = -sinx => f''(0)/2! =0
f'''(x) = -cosx => f'''(0)/3! =-1/6
sinx ~ x - (1/6)x^3
不知道怎么化解啊,能详细说一下嘛?我以为要用到洛必达法则的0/0型呢