High number problem: when x tends to 0, if the formula of X contains 1 / x, can't we use the formula of equivalent infinitesimal? When x tends to 0, if the expression of X contains 1 / x, can we not use the formula of equivalent infinitesimal? When x tends to infinity, if the expression of X contains 1 / x, can we use the formula of equivalent infinitesimal? For example, e ^ [(1 / (x (x-1))] - 1

High number problem: when x tends to 0, if the formula of X contains 1 / x, can't we use the formula of equivalent infinitesimal? When x tends to 0, if the expression of X contains 1 / x, can we not use the formula of equivalent infinitesimal? When x tends to infinity, if the expression of X contains 1 / x, can we use the formula of equivalent infinitesimal? For example, e ^ [(1 / (x (x-1))] - 1

Not because of this, but because to make sure that the whole is infinitesimal, for example, let e ^ [(1 / (x (x-1))] - 1 be infinitesimal, 1 / (x (x-1)) should be infinitesimal, but actually 1 / (x (x-1)) is infinite, and the whole should also be infinitesimal, not infinitesimal at all. If e ^ [(SiNx ^ 4 / (x (x-1))] - 1 can be e ^ [(SiNx ^ 4 /...)

SiNx * e ^ x primitive function

F(x) = (sinx.e^x - cosx.e^x)/2 + c
F'(x) = [( cosx.e^x + sinxe^x )- (-sinx.e^x +cosx.e^x)]/2
= sinx .e^x

When x tends to zero, the order of the infinitesimal e ^ x + SiNx - 1 with respect to the basic infinitesimal x is determined

It is to find LIM (x approaches 0) {[e ^ x + sinx-1] / X}
You can use the lobita rule
By deriving the denominator of {[e ^ x + sinx-1] / X}, we get {[e ^ x + cosx]} / 1
When x approaches 0, we get 1 + 1 = 2, so the order of the infinitesimal e ^ x + sinx-1 with respect to the basic infinitesimal x is the infinitesimal of the same order