设函数f(x)在x=1处连续,且limf(x)/x-1=2,则f(1)等于A,-1 B,0 C,1 D,2详细的过程,谢谢

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0由于连续,所以在此点limf(x)=f(x)所以limf(x)/(x-1)=f(x)/(x-1)=2所以f(x)和x-1是同阶无穷小。当然等于0或者说f(x)=(1-1)*2=0