HMMT 二月 2008 · CALC 赛 · 第 4 题

HMMT February 2008 — CALC Round — Problem 4

[ 4 ] Let a, b be constants such that lim = 1. Determine the pair ( a, b ). 2 x → 1 x + ax + b ( ) ( ) 6 x x 6 (2008)

解析

[ 4 ] Let a, b be constants such that lim = 1. Determine the pair ( a, b ). 2 x → 1 x + ax + b Answer: ( − 2 , 1) When x = 1, the numerator is 0, so the denominator must be zero as well, so 1 + a + b = 0. Using l’Hˆ opital’s rule, we must have 2 (ln(2 − x )) 2 ln(2 − x ) 1 = lim = lim , 2 x → 1 x → 1 x + ax + b ( x − 2)(2 x + a ) and by the same argument we find that 2 + a = 0. Thus, a = − 2 and b = 1. This is indeed a solution, as can be seen by finishing the computation. ( ) ( ) 6 x x 6 (2008)