Suppose that function f is differentiable. Prove that a given limit exists and equals f '(x)

(a) lim(h-->0) [2 f(x+h) - f(x) - f(x-h)]/3h

Since f is differentiable, you can use L'Hopital's Rule.

lim(h-->0) [2 f(x+h) - f(x) - f(x-h)]/(3h)
= lim(h-->0) [2 f '(x+h) - 0 - (-f '(x-h))] / 3, differentiating with respect to h
= [2 f '(x) - 0 + f '(x)] / 3
= f '(x).

I hope this helps!