Limit problems related to Taylor formula Find limit Lim [x-x ^ 2ln (1 + 1 / x)] (x → + ∞)

Limit problems related to Taylor formula Find limit Lim [x-x ^ 2ln (1 + 1 / x)] (x → + ∞)

When X - > positive infinity, 1 / X - > 0, there is
ln（1＋1/x）＝1/x－1/（2x^2）+1/(3x^3)-1/(4x^4)+0(1/(x^4))
therefore
The original formula = Lim [x-x + 1 / 2-1 / (3x) + 1 / (4x ^ 2) - x ^ 2 * 0 (1 / x ^ 4)] = 1 / 2
I don't have a formula editor, where 0 () denotes low order infinitesimal,

Taylor's formula for limit
The limit of (x ^ 3 + 3 * x ^ 2) ^ 1 / 3 - (x ^ 4-2 * x ^ 3) ^ 1 / 4 when X - > ∞
Please tell me how to use Taylor formula to find the limit of this formula. In particular, I want to ask, after the derivation, the denominator part of the derivative can't be brought into 0, so McLaughlin formula can't be applied. How to solve this situation? Is there any way to simplify the calculation process for this type of derivation

∵ (1 + x) ^ α = 1 + α x + α (α - 1) (X & # 178 / 2) + O (X & # 178;) (Taylor formula, O (x) is infinitesimal of higher order) ∵ (X & # 179; + 3x & # 178;) ^ (1 / 3) = x (1 + 3 / x) ^ (1 / 3) = x [1 + (1 / 3) (3 / x) + (1 / 3) (1 / 3-1) ((3 / x) & # 178 / 2) + O (1 / X & # 178;)] (expand Taylor formula) = x [1

A proof of Taylor formula
Let f (x) have n + 1 order continuous derivative near the point, and f '(x0) = f' '(x0) =... = FN (x0) = 0, f (n + 1) (x0) ≠ 0. It is proved that if n is odd, then point x0 is the extreme point of F (x); if n is even, then point x0 is not the extreme point of F (x)

For the Taylor formula of F (x) at x0, because f '(x0) = f' '(x0) =... = FN (x0) = 0, the second term to the nth term in the Taylor formula are all 0, so only the first term and the nth + 1 term are left, that is, f (x) = f (x0) + [f (n + 1) (x0) / (n + 1)!] (x-x0) ^ (n + 1), so the left and right derivatives of the formula are f' (x) = [f (n + 1) (x