When x tends to zero, which is higher order infinitesimal compared with SiNx (TaNx + x ^ 2)

SiNx (TaNx + x ^ 2) ~ x * TaNx ~ x * x = x ^ 2 (when X - > 0)
So SiNx (TaNx + x ^ 2) is infinitesimal of higher order

How many infinitesimals is f (x) = (1 + x ^ 2 / 2) cos X-1
Why does cos x need to be replaced by the fifth order infinitesimal of 1-x ^ 2 / 2 + x ^ 4 / 24 + X instead of the third order infinitesimal of 1-x ^ 2 / 2 + X or the first order infinitesimal of 1 + X? For example, I use the third order infinitesimal of 1-x ^ 2 / 2 + X to get 0 (x ^ 3). Why can't I say that f (x) is the third order infinitesimal? Or I use the first order infinitesimal of 1 + X to get the first order infinitesimal of x ^ 2 / 2 + X, Why can't we say that f (x) is an infinitesimal of order 1?

This is called loss of rank~
The Taylor expansion of COS x is even, you remember~
Turn over the book:

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