f(0)=0,f'(0)=2,则lim(x~0)f(1—cosx)/x^2=?.
已知lim(x→0) f(x)/(1-cosx) =2 求lim(x→0) [1+f(x)]^½
设f(x)具有一阶连续导数,f(0)=0,f'(0)=2,求了lim(x→0)f(1-cosx)/tan(x^2)
已知lim(x→0) [f(0)-f(2x)]/x=1,求f'(0).
f(x)有定义,f(2x)=f(x)cos x,lim f(x)=f(0)=1(x趋于0时),求f(x)
lim(x趋向于0)f(2x)/x=1,且f(x)连续,则f'(0)=
f(x)在x=a处有二阶导数,求证x趋于0时lim(((f(a+x)-f(a)/x}-f‘(a))/x=1/2f''(a
设函数f(x)连续,lim((f(x)/x)-1/x-(sinx/x^2))=2,f(0)=?
大学高数极限求解已知lim┬(x→0)〖(x^2 f(x)+cosx-1)/x^4 〗=0 求lim┬(x
求解一道极限运算题lim{sin2x+xf(x)}/x^3=1 (x→0) lim{2cosx+f(x)}/x^2这类题
设函数 F(x)在x=0处可导 又F(0)=0,求lim(x→0) F(1-cosx)/tan(x²)
设f(x)导数在【-1,1】上连续,且f(0)=1,计算∫【f(cosx)cosx-f‘(cosx)sin^2x】dx(
lim x趋于0 f(x)/x^2=5 求lim x趋于0 f(x)=?