 # How to prove that when X - > 0, X is equivalent to arctanx

## How to prove that when X - > 0, X is equivalent to arctanx

Using lobida's law
lim arctanx/x
=lim 1/(1+x^2)
=1
So when x → 0, arctanx ~ X

### Equivalent infinitesimal of arctanx

X when x tends to 0

### Tongji high numbers Volume 1: the question of the derivation of the derivative of power function using the definition of derivative? (X^n-A^n)/(X-A)==?

X^n-A^n=A^n((X/A)^n-1)=A^n(X/A-1)((X/A)^(n-1)+(X/A)^(n-2)+……+X/A+1)
=(X-A)(X^(n-1)+AX^(n-2)+A^2*X^(n-3)+……+A^(n-2)*X+A^(n-1))
∴(X^n-A^n)/(X-A)=X^(n-1)+AX^(n-2)+A^2*X^(n-3)+……+A^(n-2)*X+A^(n-1)

### Comparison and inquiry of Infinitesimal in Tongji advanced mathematics sixth edition Dear everyone... When I saw the sixth edition p57 of Tongji advanced mathematics, I had a big head because I couldn't understand what this thing was: If Lim（ β/α)= 0 just say β Yes than α The infinitesimal of higher order is recorded as β= o( α) So the problem arises, I can't understand β= o( α) It's a divine horse... Turn to the next page, P58, theorem 1 on equivalent infinitesimal: β And α Is an equivalent infinitesimal. The necessary and sufficient conditions are: β=α+ o( α) I don't understand My personal idea 1: β= o( α) Yes α To represent β A function of feels that this is wrong Personal idea 2: Equivalent time β=α+ o( α)? When explicitly equivalent β=α Well, is that o（ α)= 0 I can't understand it at all. Ask for help

1、 α Is infinitesimal, and α Same, O（ α) It is also an infinitesimal, but this infinitesimal ratio α Higher order, that is, LIM (o（ α)/α)= 0.o( α) It's just a representation of those ratios α Higher order infinitesimal, α And β (i.e. o（ α)） There is no functional relationship between them; 2. To correct one of your errors

Lobida's law
Or expand e ^ X

### High number equivalent infinitesimal problem (can the function in the function be equivalent to infinitesimal) For example, when x tends to 0, find the limit of Ln (tan2x) / ln (tan7x). Can I first reduce the equivalent infinity in brackets to ln (2x) / ln (7x), and then lobita. For this example, the answer is 1 is correct. I want to know whether the function in any function can be equivalent to infinitesimal (note that it is not the multiplication of two functions, but in the function) , if not, can you give a counterexample. Yes, give the reason. Note: what I ask is different from this kind. When x tends to 0, sin (sin2x) is equivalent to 2x, because this kind of function is from outside to inside, and what I ask is from inside to outside. I wonder if you can understand what I ask. Hehe

On the problem of Equivalent Infinitesimal Substitution, don't recite the conclusion. You should know the principle, especially why it is right when you do it right, otherwise it's no different from what you guessed right
For the specific questions you give, pay attention to when X - > 0 +
lim ln(tan2x)/ln(2x) = 1 + lim [ln(tan2x)-ln(2x)]/ln(2x) = 1
So it can lead to the substitution of equivalent infinitesimal
Of course, I don't think such a replacement is of any value. The difficulty of proving that it can be replaced is the same as that of the original problem, but it is convenient for you to use the L'Hospital law, but such problems can be solved without using the L'Hospital law at all
Abstract your question again. Under a certain change trend (such as X - > A), Lim f (x) / g (x) = 1, and H (x) has certain continuity, can Lim H (f (x)) / h (g (x)) = 1 also be guaranteed?
Generally speaking, the conclusion is wrong. Let me give you a counterexample:
When X - > 0, f (x) = 1 / x ^ 4, G (x) = 1 / x ^ 4 + 1 / x ^ 2, H (x) = e ^ X
If you have to be infinitesimal instead of infinitesimal, for example
When X - > 0, f (x) = x ^ 2, G (x) = x ^ 2 + x ^ 4, H (x) = e ^ {- 1 / x ^ 2}