设f(x)在x=1处连续,且lim(x→1)[f(x)/(x-1)]=2,求f`(1)=?我想知道得出f(1）=0的详细过程

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). 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=1处连续,且lim(x趋向于1时)f(x)/(x-1)=2,则f'(1)=___ 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 设函数f(x)在x=2处连续,且lim(x)/(x-2)(x→2)=2,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 有关极限,设函数f(x)在x=2处连续,且f(2)=3,求lim(x→2)f(x)=[(1/x-2)-(4/x^2-4)] 设f(x)在x=1处连续,且lim(x→1)[f(x)/(x-1)]=2,求f`(1)=?我想知道得出f(1）=0的详细过程 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)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 （f（x）-1）/x=-1,x→0.,求f(0)期待更详细的回答~为什么否则已知的极限不存在？ 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)有二姐连续导数,且(x->0)lim[f(x)-a]/[e^x^2-1]=0,(x->0)lim[f ‘’(x)+1]/[1-cosx]=2,则答案为f(x)在x=0处取极大值李永乐复习全书p95解答上有一步 (x->0)lim[f''(x)+1]=lim[f''(x)+1]/0.5*x^2=2 ,由此可知 lim[f''( 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)=∞ 已知函数f(x)在x＝1处连续,且lim(f(x)/(x－1)²)＝2,求f'(1)x→1 设f (x)在x=0处可导,且f (0)=0,求证：lim(x→∞)f (tx)-f (x)/x=(t-1)f' (0)