Let R be the reals. Let f : (-1, 1) → R be a function with a derivative at 0. Let an be a sequence in (-1, 0) tending to zero and bn a sequence in (0, 1) tending to zero. Show that limn→∞ (f(bn) - f(an)/(bn - an) = f '(0).
Solution
From the definition of derivative, we have that ( f(bn) - f(0) )/bn → f '(0), and ( f(0) - f(an) )/-an → f '(0). So f(bn) - f(0) lies within ε/2 of bnf '(0) for sufficiently large n, and f(0) - f(an) lies within ε/2 of -anf '(0) for sufficiently large n. Hence, adding, f(bn) - f(an) lies within ε of (bn - an)f '(0) for sufficiently large n, which is what we need.
© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999