若f'(1)=3,则lim_(h->0)(f(1)-f(1-2h))/h= ( )
求解两道极限题若函数f(x)再a点可导,则lim_[f(a)-f(a+2h)]/3h=?设f(x)在x=0处连续,则li
设f(x)在x=2处可导,且f'(2)=1,则lim h→0 [ f(2+h)-f(2-h)]/h等于多少,
高数,求极限若f'(x0)=1,则lim h→0 = [ f(x0+2h)-f(x0) ] / h若f'(x0)=1,则
设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0
f(0)=0,lim [f(1-cos h)/(h^2)](h->0)存在,能否得到f'(0)存在
设函数f(x)在x=1处可导,且f'(1)=2,则[lim(h→0)f(1-h)-f(1)]/h等于
已知函数f(x)在x0可导,且lim(h→0)h/[f(x0-2h)-f(x0)]=1/4,则f‘(x0)=?
若f′(x0)=-2,则lim[f(x0+h)-f(x0-h)]/h=
设f'(x) = 3^(1/2) ,求 lim(h→0) [f(x+mh) - f(x - nh)] / h ,(m ,
若函数f(x)在点x=a处可导,则lim(h→0)[f(a+4h)-f(a-2h)]/3h=?
f(x)=sinx,求{f(1+h)-f(1)}/h 结果{2sin h/2 cos 2+h/2}/h是如何求出来的
设函数f(x)在x=a处可导,且lim[f(a+5h)]-f(a-5h)]/2h=1,则f'(a)=