When x tends to 0, both up and down are 0, and the amount is infinitesimal.

So the original formula =lim (x tends to 0)[f(2+x)-f(2-x)]'/x'

=lim (x tends to 0)[f'(2+x)-f'(2-x)*(-x)']

=lim (x tends to 0)[f'(2+x)+f'(2-x)]

=f'(2)+f'(2)=6