When x → 0, (1-ax ^ 2) ^ (1 / 4) - 1 and xsinx are equivalent infinitesimals | Math enthusiast | Mathematical art

When x → 0, (1-ax ^ 2) ^ (1 / 4) - 1 and xsinx are equivalent infinitesimals

SiNx is equivalent to X when x tends to 0. Xsinx is equivalent to X & # 178;. Then (1-ax & # 178;) ^ (1 / 4) - 1 is regarded as a function of F (x), and its expansion at 0 point is: F (0) + X * f '(0) + X & # 178; f (θ) / 2!. this is Taylor expansion. Let's see f' (x) = (1 / 4) (1-ax & # 178;) ^ (1 / 4-1) (- 2aX) = (- ax / 2) (1-A

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