求lim【h→0】1/h∫【x-h→x+h】cost^2dt(h>0)
求极限!分母为零!像lim h→0 ((1+h)^(1/2)-1)/h和lim x→-1 (x^2+2x+1)/(x^4
f(x)在x=a处可导, lim(h→0) [f(a+h)-f(a-2h)]/h=
f(x)在x_0处可导,求lim h→0 f(x_0+h)-f(x_0-h)/5h 的值
设函数f(x)在点x处可导,试求h→0 lim f(x+h)-f(x-h)/2h的值
设f'(x) = 3^(1/2) ,求 lim(h→0) [f(x+mh) - f(x - nh)] / h ,(m ,
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,
求解一道高数极限题lim[tan(x+h)-2tanx+tan(x-h)]/h² h→0 答案为2sec
设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值
设函数f(x)在[A,B]上连续,证明lim(h→0) 1/h*∫(x,a)[f(t+h)-f(t)]dt=f(x)-f
若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2.
lim[tan(π/3+h)-tan(π/3-h)]/h h→0求极限
其请问 lim(h→0) [ f(x0+3h)-f(x0-2h) ] / h