This is to use the relationship between limit and infinitesimal to find limit If LIM (x → 0) (sin6x + XF (x)) / x ^ 3 = O, then LIM (x → 0) (6 + F (x)) / x ^ 2 is But what I calculate is 0 for LIM (x → 0) (sin6x + XF (x)) / x ^ 3 = O. if the left side of the equation is divided by X, we will get LIM (x → 0) (6 + F (x)) / x ^ 2 = O. the error in solving the equation is

This is to use the relationship between limit and infinitesimal to find limit If LIM (x → 0) (sin6x + XF (x)) / x ^ 3 = O, then LIM (x → 0) (6 + F (x)) / x ^ 2 is But what I calculate is 0 for LIM (x → 0) (sin6x + XF (x)) / x ^ 3 = O. if the left side of the equation is divided by X, we will get LIM (x → 0) (6 + F (x)) / x ^ 2 = O. the error in solving the equation is

You have directly changed sin6x / x into 6. The reason why I use "change" instead of "seek" is that sin6x / X is equal to 6 only after limit operation, and the sum formula sin6x / x + F (x) is a fraction molecule. Since molecules can seek limit, we can't only seek limit for a part of sin6x / X, Does f (x) not seek the limit? Does the denominator not seek the limit? That is to say, it cannot be Lim [((LIM sin6x / x) + F (x)) / x ^ 3]. No limit algorithm or formula can do this
The correct way is to find out the formula to calculate the limit, that is, (sin6x + XF (x)) / x ^ 3 = (6x + XF (x)) / x ^ 3 + (sin6x-6x) / x ^ 3 = (6 + F (x)) / x ^ 2 + (sin6x-6x) / x ^ 3. As long as the limit of (sin6x-6x) / x ^ 3 is calculated, LIM (6 + F (x)) / x ^ 2 can be obtained according to the sum difference algorithm of the limit