functions and parameter

Question

I have the function $f(x)=|\ln{x}-ax+a|,x>0,a\in\mathbb{R}$ which can derives and $x_0=1$. Now I want to prove firstly that $f'(1)=0$ and secondly that $a=1$. I am thinking of using that:$$f'(x_0)=\lim_{x\to1^-}{\frac{f(x)-f(x_0)}{x-x_0}}=\lim_{x\to1+}{\frac{f(x)-f(x_0)}{x-x_0}}$$ But this didn't waork for me and I have been confused. Any help?

Answer 1

Use that $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=\frac{|\ln(1+h)-a(1+h)-a|-|\ln(1)-a+a|}{h}$$