若f(x)可导,f'(x)=0,lim(x趋于0)f'(x)/x=-1,求证:f(0)是极大值.

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 已知函数f(x)可导 且lim(x无限趋于0)f(1)-f(1-x) /x=-1 则f'(1)= 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) 若f(x)与g(x)可导,Lim f(x)=Limg(x)=0,且Limf(x)/g(x)=A,x趋于a.则 lim x趋于0 f(x)/x^2=5 求lim x趋于0 f(x)=? 设f(x)有二阶导数,且f''(X)>0,lim(x趋于0）f(x)/x=1 ..证明：当x>0时,有f(x)>x 若f﹙x﹚在x=0的某邻域内可导,且lim x→0 f'﹙x﹚=1 则f﹙0﹚ A.是f(x)的极大值 B.是f(x)的 设f(x)为可导函数,且满足lim[f(1)+f(1-x)]/2x=-1,x趋于0时,求曲线y=f(x)在点（1,f(1 设f(x)为可导函数,且满足lim[f(1)-f(1-x)]/2x=-1,x趋于0,求曲线y=f(x)在点（1,f(1) 设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点（1,f( 设f(x)为可导函数,且满足lim[f(1)-f(1-x)]/2x=-2,x趋于0时,求曲线y=f(x)在点（1,f(1 设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(