Your last 3 steps should not have the limit statement in front of them because you already took the limit. The next to last step to the last step is wrong:
$\frac{4-2a}{2-a} = \frac{2(2-a)}{2-a} = 2$
It would be much better for you to type the equations so we could edit your post.
$g(x)=2x-1$
derivative of $g$ at any value $x=a$ in the domain of $g$
$\displaystyle g’(a) = \lim_{x \to a} \dfrac{g(x)-g(a)}{x-a}$
$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2x-1 - (2a-1)}{x-a}$
$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2x-2a}{x-a}$
$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2(x-a)}{x-a}$
$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2( \cancel {x-a})}{\cancel {x-a}} = 2$