f（x）为连续函数,x趋于2时,［f（x）+3］/ln（x-1）=1,y=f（x）在x=2处的切线方程为?

∵lim(x→2) ln(x-1) =0∴lim(x→2) [f(x)+3]=0, 即 f(2)=-3lim(x→2) [f(x)+3]/ln(x-1)=lim(x→2) [f(x)+3]'/[ln(x-1)]'=lim(x→2) [(x-1)f'(x)]=f'(2)=1切线方程为y-f(2)=f'(2)*(x-2),即y=x-5