Hello

$\displaystyle \frac{f(a+h)-f(a)}{h}=\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac 12}{h} \quad (*)$

now, by noting that 2(h+2) is the common denominator, $\displaystyle \frac{1}{2+h}-\frac 12=\frac{2}{2(2+h)}-\frac{2+h}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}$

$\displaystyle (*)=\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2h(2+h)}=-\frac{1}{2(2+h)}$

now take the limit h -> 0 and you'll have the answer

There is no difference

$\displaystyle f'(a)=\lim_{h \to 0} ~ \frac{f(a+h)-f(a)}{h}=\lim_{x \to a} ~ \frac{f(x)-f(a)}{x-a}$

from the first one to the second one, you just have to substitute $\displaystyle x=a+h$