Putnam 1954 B5 Session B Let f(x)f(x)f(x) be a real-valued function, defined for −1<x<1-1<x<1−1<x<1 for which f′(0)f'(0)f′(0) exists. Let (an),(bn)(a_n) , (b_n)(an),(bn) be two sequences such that −1<an<0<bn<1-1 <a_n <0 <b_n <1−1<an<0<bn<1 for all nnn and limn→∞an=0=limn→∞bn.\lim_{n \to \infty } a_n = 0 =\lim_{n \to \infty} b_n.limn→∞an=0=limn→∞bn. Prove that limn→∞f(bn)−f(an)bn−an=f′(0).\lim_{n \to \infty} \frac{ f(b_n )- f(a_n ) }{b_n -a_n} =f'(0).n→∞limbn−anf(bn)−f(an)=f′(0). Dataset text AI-MO AoPS. Quoted with source. Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution Similar problems These problems have the most words in common with this statement. TM-MO 2015 Problem 2Putnam 2021 A2Putnam 1970 A4Putnam 1947 A1Putnam 1941 B2 PreviousB4 Locator putnam/1954/b5 · Tag 94MFEJVCKT · Edit this page on GitHub · Keys: j next, k previous