Given LIM (x - > 0) (2arctanx ln (1 + X / 1-x)) / x ^ n = C! = 0, find the values of constants C and N. the solution of this problem is done by using the law of Robida, My method is to split (2arctanx ln (1 + X / 1-x)) / x ^ n into (2arctanx / x ^ n) - (LN (1 + X / 1-x)) / x ^ n, and then use the equivalent infinitesimal to solve it. Why is the result different from the original solution? When is the appropriate time to use the equivalent infinitesimal

Given LIM (x - > 0) (2arctanx ln (1 + X / 1-x)) / x ^ n = C! = 0, find the values of constants C and N. the solution of this problem is done by using the law of Robida, My method is to split (2arctanx ln (1 + X / 1-x)) / x ^ n into (2arctanx / x ^ n) - (LN (1 + X / 1-x)) / x ^ n, and then use the equivalent infinitesimal to solve it. Why is the result different from the original solution? When is the appropriate time to use the equivalent infinitesimal

If it is an infinitive of 0 / 0, ∞ / ∞ or 0 ·∞ and there is a single replaceable equivalent infinitesimal factor in the whole formula, then it can be replaced by a simpler factor
However, if the molecules are separated by + or - we should be careful not to split the whole formula by + - sign
If there are two limits after disassembly, it can be disassembled

How to calculate the limit of (secx-1) / (x ^ 2) when x → 0 is 1 / 2?
Please give me your advice

The first method: lobida's law, which is of type 0 / 0, obviously satisfies the condition of the theorem. By using the upper and lower derivatives of the theorem twice, the result can be obtained
The second method: equivalent infinitesimal: for example, cosx is equivalent to 1-x ^ 2 / 2, SiNx is equivalent to x, e ^ X-1 is equivalent to x,
When x tends to 0
The third method is still equivalent infinitesimal, but it is based on Taylor expansion, which is the theoretical basis of the second method

The limit of (secx cosx) / (x ^ 2)