Could any body please give me a hint about the first step
Find $\displaystyle \lim_{x \to 2} \left[ \frac{x-1}{x-2} - \frac{1}{ln(x-1)} \right]$
The difficulty with a question like this, where you show no work at all, is that we have no idea what you have available. Since this is of the form "$\displaystyle \infty- \infty$", I would be inclined to combine the fractions, changing it into a "0/0" situation, then use L'Hopital's rule.
just as what HallsofIvy said, next time you should show your effort in your post, then we can help you in a better and more efficient way.
$\displaystyle \lim_{x\to 2}(\frac{x-1}{x-2}-\frac{1}{ln(x-1)})
=\lim_{x\to 2}\frac{(x-1)ln(x-1)-x+2}{(x-2)ln(x-1)}$
$\displaystyle =\lim_{x\to 2}\frac{ln(x-1)+1-1}{ln(x-1)+\frac{x-2}{x-1}}$ (Hopital's rule)
$\displaystyle =\lim_{x\to 2}\frac{(x-1)ln(x-1)}{(x-1)ln(x-1)+x-2}$
$\displaystyle =\lim_{x\to 2}\frac{ln(x-1)+1}{ln(x-1)+1+1}$ (Hopital's rule again!)
$\displaystyle =\frac{1}{2}$.