Why does limx tend to 1, ln (1 + 2x) / X-1 = infinite? Why not equal to 2x / X-1 = 2?

Why does limx tend to 1, ln (1 + 2x) / X-1 = infinite? Why not equal to 2x / X-1 = 2?

lim(x->1) ln(1+2x) = ln3
lim(x->1)(x-1) =0
LIM (x - > 1) / (x-1) = infinite
Ln (1 + x) is equivalent to X when X - > 0
It's not X - > 1

Limx tends to zero. What is the negative x ^ 2 + 2x / absolute value x (x + 1)

Limx tends to 0 (x (2-x) / lxllx + 1L
x>0
Limx tends to 0 (x (2-x) / lxllx + 1L = 2
x

Can LIM (LN (1 + 1 / x) / arccotx) replace ln (1 + 1 / x) with (1 + x) by equivalent infinitesimal

You can't
Only when x →∞, ln (1 + 1 / x) can be replaced by 1 / X

Can Lim X -- > 0 be used as the equivalent infinitesimal? Can Lim X -- > ln (1 + 1 / x) / X be replaced by 1 / x

For example
ln(1+x)~x
This does require x → 0
However, the actual application of
We can see it as
ln(1+□)~□ (□→0)
Therefore, the situation in your question can be applied

The difference between basic limit and equivalent infinitesimal?
In the form of the formula, infinitesimal is a few more than the basic limit (such as Tan, arctan, etc.). Is everything else the same?
Is there any difference in usage?
Isn't it all equivalent substitution?

Infinitesimal and limit are two different concepts. You should pay attention to distinguish them, but I don't quite understand what you mean. The definition of infinitesimal is a variable whose limit is the number zero. To be exact, when the independent variable x is infinitely close to x0 (or the absolute value of X is infinitely increased), the function value f (x) is infinitely close to zero, that is, f (x) = 0 (or F (x) = 0), then f (...)

When x approaches 0, ln (1 + 2xarcsinx) / Tan ^ 2x limit

When x approaches 0, ln (1 + 2xarcsinx) / Tan ^ 2x limit
=lim(x->0) 2xarcsinx/(x^2)
=lim(x->0) 2x^2/(x^2)
=2