Differentiation Question 2

Idea of the proof of Thm 4.3.1????

2011-11-10 22:51:32 補充：
(1) Since lim(h->0-) f(x0+h) = f(x0)
so, for any ε>0, there exists δ>0, such that,
| f(t) - f(x0) | < ε, whenever t in (x0-δ, x0).
Taking h=-δ and t in (x0+h, x0), then | f(t)- f(x0) | < ε
so,
(1/h)∫_[x0, x0+h ] f(t) dt - f(x0)
=(-1/h)∫_[x0+h, x0] [ f(t)- f(x0)] dt (since (-1/h)∫_[x0+h, x0] f(x0)dt= f(x0) )
then
| (1/h)∫_[x0, x0+h] f(t) dt - f(x0) |
<= (-1/h)∫_[x0+h, x0] | f(t)- f(x0) | dt
< (-1/h) ε*(-h)
=ε
Hence, lim(h->0-) (1/h)∫_[x0, x0+h] f(t) dt = f(x0)

(2) Since lim(h->0+) f(x0+h) = f(x0)
so, for any ε>0, there exists δ>0, such that,
| f(t) - f(x0) | < ε, whenever t in (x0, x0+δ).
Taking h=δ and t in (x0, x0+h), then | f(t)- f(x0) | < ε
so,
(1/h)∫_[x0, x0+h ] f(t) dt - f(x0)
=(1/h)∫_[x0, x0+h] [ f(t)- f(x0)] dt (since (1/h)∫_[x0, x0+h] f(x0)dt= f(x0) )
then
| (1/h)∫_[x0, x0+h] f(t) dt - f(x0) |
<= (1/h)∫_[x0, x0+h] | f(t)- f(x0) | dt
< (1/h) ε*(h)
=ε
Hence, lim(h->0+) (1/h)∫_[x0, x0+h] f(t) dt = f(x0)

(3) Since lim(h->0) f(x0+h)= f(x), then both of (1),(2) are hold.
Hence, Hence, lim(h->0) (1/h)∫_[x0, x0+h] f(t) dt = f(x0)