设f(x)在x=0连续,且lim(x+sinx)/ln[f(x)+2]=1x趋近于0,则f'(0)?

设f(x)在x=0处连续,且lim(x趋于0)f(x)/x存在,证明,f(x)在x=0处可导 证明：设f(x)在x=0连续,且lim(x→0) (f(x)/x)=1,则必有f'(0)=1 设f(x)在x=0连续,且lim(x+sinx)/ln[f(x)+2]=1x趋近于0,则f'(0)? 设函数f(x)在x=2处连续,且lim（x→2）f(x)/(x-2)(x→2)=3,求f'(2). 设函数t(x)在点X=6处连续,且f(6)= -5 则 lim f(x)=?lim是 x->6 设f(x)具有连续导数,且满足f(x)=x+∫(上x下0)tf'(x-t)dt求lim(x->-∞)f(x) 设f(x)在x=0处连续,且lim(x趋于0）f(x)/x^2=1 ,证明函数f(x)在x=0处可导且取得极小值. 设f(x)在x=0处连续,且lim(x趋于0）f(x)/x^2=1 ,证明函数f(x)在x=0处可导且取得极小值. 设f(x)在x=1处连续,且lim(x趋向于1时)f(x)/(x-1)=2,则f'(1)=___ 设f(x)连续,g(x) =∫(1,0)f(xt)dt,且lim x→0 f(x)/x =A,求 g'(x).如题 设f(x)在x=0的某一邻域内二阶可导,且lim(x-->0)f(x)/x=0,f''(0)=2.求lim(x-->0)f(x)/x^2因为f(x)在x=0处二阶可导从而连续且lim(x-->0)f(x)/x=0为什么能得到lim(x-->0)f(x)=f(0)=0.请详细解释,多谢 设f(x)在[0,+∞)上有连续的一阶导数,且lim(x→∞)f'(x)=a,证lim(x→∞)f(x)=∞ 设f(x)有二阶连续导数 且f(0)=f'(0)=0 f''(0)>0 又设u=u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴的截距则lim(x→0) x/u(x)=?求截距这个很简单了,直接就是u(x)=[xf'(x)-f(x)]/f'(x)然后我得到lim(x→0) x/u(x)=lim(x→ 设函数f(x)在(a,+∞ )上可导,且lim(x->+∞ )(f(x)+f'(x))=0,证明:lim(x->+∞ )f(x)=0 设函数f(x)在x=2处连续,且lim(x)/(x-2)(x→2)=2,求f'(2). 设f(x)有二阶连续导数且f'(x)=0,lim(x趋向于0)f''(x)/|x|=1则 设函数t(x)在点X=6处连续,且f(6)= -5 则 lim f(x)=?x->0打错了 条件是lim是 X->6的 若函数f(x)在x=0处连续,且lim{x趋近0}f(x)/x存在,试证f(x)在x=0处可导