极限题目，图中第3小题，麻烦大神帮帮解答一下吧！
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x->1/2
π-3arccosx ~ a(x-1/2)^b
By Taylor's expansion
f(x) = arccosx => f(1/2) =π/3
f'(x) = -1/√(1-x^2) =>f'(1/2)/1! = -(2/3)√3
f(x) = f(1/2) + [f'(1/2)/1!](x-1/2)+...
=π/3 -(2/3)√3 .(x- 1/2)
π-3arccosx
= π -3[π/3 -(2/3)√3 .(x- 1/2)+..]
=2√3 .(x- 1/2) +...
~2√3(x- 1/2)
π-3arccosx ~ a(x-1/2)^b
=> a=2√3 , b=1
π-3arccosx ~ a(x-1/2)^b
By Taylor's expansion
f(x) = arccosx => f(1/2) =π/3
f'(x) = -1/√(1-x^2) =>f'(1/2)/1! = -(2/3)√3
f(x) = f(1/2) + [f'(1/2)/1!](x-1/2)+...
=π/3 -(2/3)√3 .(x- 1/2)
π-3arccosx
= π -3[π/3 -(2/3)√3 .(x- 1/2)+..]
=2√3 .(x- 1/2) +...
~2√3(x- 1/2)
π-3arccosx ~ a(x-1/2)^b
=> a=2√3 , b=1
追问
a是正的诶
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