设函数f(x)在点a处可导,求下列极限:

h趋于0时,求[f(a+h)-f(a-2h)]/h的极限
3f(a)的导数
已知f(0)=0,a=0,在x趋于0时,求lim f(x)/x
=f(0)的导数

h->0
2h->0
lim(h->0)(f(a+h)-f(a))/h=f'(a)
lim(h->0)(f(a-2h)-f(a))/(-2h)=f'(a)
=>
lim(h->0)(f(a-2h)-f(a))/h=-2f'(a)
=>
lim(h->0)(f(a+h)-f(a-2h))/h
=lim(h->0)(f(a+h)-f(h))/h - lim(h->0)(f(a-2h)-f(h))/h
=f'(a) - (-2f'(a))
=3f'(a)
lim(x->0)f(x)/x=(f(x)-f(0))/(x-0)=f'(0)

[f(a+h)-f(a-2h)]/h=[f(a+h)-f(a)+f(a)-f(a-2h)]/h
=[f(a+h)-f(a)]/h+2*[f(a)-f(a-2h)]/2h
=f~(a)+2*f~(a)
=3f~(a) 其中f~(a)表示f(x)...