= limδx→0(x？ δx+xδx？ + 1/3δx？ )/δx
= x？
F' (0) = 0 f' (root number 2)=2
Second: 1) original formula =-(lim δ x→ 0 [f (x0-δ x)-f (x0)] divided by -δ x)
= - f '(x0)
2) original formula = 2 (lim δ x→ 0 [f (x0+h)-f (x0-h)] divided by 2h)
= 2 f '(x0)
Note: As long as the denominator is the difference between two independent variables in the numerator, the result is equal to the derivative.
For example, the difference between two independent variables in 2), the numerator = (x0+h)-(x0-h)=2h, so the denominator is changed to 2h, and the result is f'(x0).
Third: limx→2 f(x) divided by x-2 =2, because the denominator→0 and the result is constant, there must be f(x)→0(x→2).
That is, f(2)=0.
F' (2) = limx → 2 [f (x)-f (2)]/(x-2) = limx → 2f (x) divided by x-2 =2.