When x → 0, (e ^ x-1) sin2x is equivalent to [(1 + x) ^ 1 / 2-1)] ln (1 + 2x) But the answer is the same level but not equivalent... Is I wrong or the answer is wrong? I'm not sure about the exam review

When x → 0, (e ^ x-1) sin2x is equivalent to [(1 + x) ^ 1 / 2-1)] ln (1 + 2x) But the answer is the same level but not equivalent... Is I wrong or the answer is wrong? I'm not sure about the exam review

The former is ~ 2x ^ 2
The latter is ~ x ^ 2

How to find the order of Taylor formula or McLaughlin formula? What do you mean?
For example, there is a question about the fourth-order Maclaurin formula for e ^ (x ^ 2) and cosx, e ^ (x ^ 2) = 1 + x ^ 2 + x ^ 4 / 2! + O (x ^ 4) cos = 1-x ^ 2 / 2! + x ^ 4 / 4! + O (x ^ 5). Why is it to find so many items, and why is it the number of infinitesimals of higher order?

The expansion of a function with Taylor formula or McLaughlin formula is to replace the original function with a polynomial approximately. To replace a function with a polynomial of several degrees means to expand the function into several orders. Of course, this kind of substitution is different, so we need to add the remainder to be equal to the original function

McLaurin expansion of Taylor formula
In addition to the following expansions of power series, are there any McLaughlin expansions of Tan x, arcsin x, arccos x?

Yes, as long as it follows the general form of Maclaurin's formula
F (x) = continuous addition (n from 0 to infinity) x ^ n * f ^ (n) (0) / N! Expansion (where f ^ (n) (0) represents the value of the n-order derivative of F at 0), but the form of the final term is not regular (it also depends on the value of f ^ (n) (0))