✔ 最佳答案
definition: f ' (x) = lim(h->0) [f(x+h) - f(x)] / h
lim(h->0) [f(1 + 3h) - f(1 - 4h)] / 7h
=lim(h->0) [f(1- 4h + 7h) - f(1 - 4h)] / 7h
=lim(7h->0) { f [(1- 4h) + 7h] - f(1 - 4h) } / 7h (因為 h->0 => 7h->0)
=lim(h->0) f ' (1 - 4h)
=f ' (1)
or
這個更詳細
lim(h->0) [f(1 + 3h) - f(1 - 4h)] / 7h
=(1/7)lim(h->0) [f(1 + 3h) - f(1- 4h)] / h
=(1/7)lim(h->0) [f(1 + 3h) - f(1) + f(1) - f(1 - 4h)] / h
=(1/7)lim(h->0) [f(1 + 3h) - f(1)] / h
+ (1/7)lim(h->0) [f(1) - f(1 - 4h)] / h
=(3/7)lim(h->0) [f(1 + 3h) - f(1)] / 3h
- (4/7)lim(h->0) [f(1 - 4h) - f(1)] / 4h
=(3/7)lim(3h->0) [f(1 + 3h) - f(1)] / 3h (因為 h->0 => 3h->0)
+ (4/7)lim(-4h->0) {f(1 + (-4h) - f(1)} / (-4h) (因為 h->0 => -4h->0)
=(3/7)lim(A->0) [f(1 + A) - f(1)] / A (A=3h)
+ (4/7)lim(B->0) [f(1+B) - f(1)] / 4h (B=-4h)
=(3/7)f ' (1) + (4/7)f ' (1)
=f ' (1)
所以lim(h->0) [f(1 + 3h) - f(1 - 4h)] / 7h = f ' (1)
而且lim(h->0) [f(1 + 3h) - f(1 - 4h)] / h = (1/7)f ' (1)