Can f (x) = g (x) + O (x) (higher order infinitesimal) be discarded in calculation? Another is the formula of Taylor series. As long as there is a point x0 in the interval (a, b) with n-order derivative, it can be transformed into the form of Taylor series. Does the whole interval (a, b) have n-order derivative? The content of this picture is very little, which is hard to understand, The problem I encountered is whether the higher order infinity of F (x) - f (x0) = the second derivative of F (x0) * (x-x0) ^ 2 + O ((x-x0) ^ 2) can be omitted Let me comment:

Can f (x) = g (x) + O (x) (higher order infinitesimal) be discarded in calculation? Another is the formula of Taylor series. As long as there is a point x0 in the interval (a, b) with n-order derivative, it can be transformed into the form of Taylor series. Does the whole interval (a, b) have n-order derivative? The content of this picture is very little, which is hard to understand, The problem I encountered is whether the higher order infinity of F (x) - f (x0) = the second derivative of F (x0) * (x-x0) ^ 2 + O ((x-x0) ^ 2) can be omitted Let me comment:

Using the Taylor expansion of the small o remainder, as long as f has n-order derivative at x0, it can be expanded to n-degree polynomial. As for when to omit the high-order infinitesimal, it depends on the specific problem. In short, it will not affect the calculation of limit after omitting, which depends on how well you master the polynomial. For example, the degree of denominator is

The substitution of infinitesimal and the omission of higher order infinitesimal in limit
Lim (1/x^2-cot^2x)=lim (1/x^2-1/tan^2x)=lim(tan^2x-x^2)/x^2*tan^2x=lim(tan^2x-x^2)/x^4
=LIM (Tan ^ 2x / x ^ 4) = lim2tanxsec ^ 2x / 4x ^ 3 = lim2xsec ^ 2x / 4x ^ 3 = limsec ^ 2 / 2x ^ 2 = lim2sec ^ 2xtanx / 4x = limsec ^ 2x * x / 2x = limsec ^ 2x / 2 = 1 / 2 (x tends to 0)
Because x ^ 2 is an infinitesimal of higher order, it is omitted

From LIM (Tan ^ 2x-x ^ 2) / x ^ 4
=LIM (Tan ^ 2x / x ^ 4) is not right
How can tan ^ 2x-x ^ 2 of the same order be omitted

Limit algorithm and Infinitesimal Substitution
limx->0 (sinx^2/x^2)/[(1-cosx)/(x^2)+(sin/x)]=1/0.5+1=4/3
The numerator and denominator are brought into sin ~ x, 1-cosx ~ x ^ 2 / 2 by equivalent infinitesimal, respectively
Analysis: however, this obviously violates the principle of Infinitesimal Substitution when adding and subtracting
The only explanation is that we use four operations of limit, and regard the above formula as three independent functions
The limit is further reduced to infinitesimal
The result should be two-thirds
Clerical error

I think the reason why he did this may be that he has made it clear that the limit of numerator denominator exists and is not zero or infinite
Then, according to the limit algorithm, the numerator and denominator can be divided after their respective limits
The condition of the molecule is established
When finding the limit of denominator, it is obvious that the limit of two terms exists and is not zero or infinite
In the same way, the four operations of limit also become the sum of their respective limits
If I write like this, surely you can understand?
However, this is only applicable when the limit energy is determined. If the limit of the denominator of a limit molecule is not known, then it can not be done, and can only be done by other methods
Therefore, this problem does not violate the principle of Infinitesimal Substitution when adding and subtracting