When x → 0, 1-cosx and axsinx are equivalent infinitesimals, then a=

1=lim(x→0)(1-cosx)/(axsinx)
=lim(x→0)(x^2/2)/(ax^2)=1/(2a)
So: a = 1 / 2

RELATED INFORMATIONS

1. When x → 0, the following functions are infinitesimal: A: SiNx / x, B: x ^ 2 + SiNx, C: ln (1 + x) / x, D: 2x-1
2. In using the Equivalent Infinitesimal Substitution to find the limit, 1: when x tends to 0, sin (f (x)) ~ f (x) 2: when f (x) tends to 0, sin (f (x)) ~ f (x) which is correct
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14. Can we use the Equivalent Infinitesimal Substitution when the limit is exponentially calculated? For example, Lim [x - > 0, (1 + SiNx) ^ (1 / 2x)], can X in index 1 / 2x be directly replaced by SiNx? Can't SiNx be directly replaced by X?
15. F (x) is the fifth order infinitesimal of X?
16. The order of 1-x and 1 / 2 (1-x ^ 2) infinitesimal when x tends to 1
17. Why is f (x) = LIM (n →∞) (1-x ^ 2n) / (1 + x ^ 2n) x equal to - x when | x | 1? Is the independent variable x? What is n?
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