LIM (arctan x-x) / SiNx ^ 3 x tends to 0, can we use the substitution of equivalent infinitesimal to solve The answer is - 1 / 3, according to the law of lobita In addition, I use the Equivalent Infinitesimal Substitution, the result is - 1 / 6 =(sin x-x) / SiN x ^ 3 x tends to zero =(cosx-1) / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) x ^ 2 / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) / 3cosx ^ 3 x tends to zero =-1/6

LIM (arctan x-x) / SiNx ^ 3 x tends to 0, can we use the substitution of equivalent infinitesimal to solve The answer is - 1 / 3, according to the law of lobita In addition, I use the Equivalent Infinitesimal Substitution, the result is - 1 / 6 =(sin x-x) / SiN x ^ 3 x tends to zero =(cosx-1) / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) x ^ 2 / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) / 3cosx ^ 3 x tends to zero =-1/6

The answer is - 1 / 6, which is also the result of lobita's law
=(cosx-1) / (cosx ^ 3) (3x ^ 2) bring x = 0 into cosx ^ 3}
=-sinx/6x
=-1/6