高数极限问题求解，看看这一步是怎么算出来的 5
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let
f(x) = (1+ x)^(1/3) =>f(0)=1
f'(x) =(1/3)(1+x)^(-2/3) =>f'(0) = 1/3
(1+x)^(1/3) ~ f(0) + f'(0) x = 1 + (1/3)x
(1+x^2)^(1/3) ~ 1 + (1/3)x^2
(1+x^2)^(1/3) - 1~ (1/3)x^2
f(x) = (1+ x)^(1/3) =>f(0)=1
f'(x) =(1/3)(1+x)^(-2/3) =>f'(0) = 1/3
(1+x)^(1/3) ~ f(0) + f'(0) x = 1 + (1/3)x
(1+x^2)^(1/3) ~ 1 + (1/3)x^2
(1+x^2)^(1/3) - 1~ (1/3)x^2
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