EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If d is any positive number and |x| ≤ d, then |f (n + 1)(x)| = ≤ 9ed. So Taylor's Inequality, with a = 0 and M = 9ed, says that |Rn(x)| ≤ (n + 1)! |x|n + 1 for |x| ≤ d. Notice that the same constant M = 9ed works for every value of n. But, from this equation, we have lim n → [infinity] 9ed (n + 1)! |x|n + 1 = 9ed lim n → [infinity] |x|n + 1 (n + 1)! = . It follows from the Squeeze Theorem that lim n → [infinity] |Rn(x)| = 0 and therefore lim n → [infinity] Rn(x) = for all values of x. By this theorem, 9ex is equal to the sum of its Maclaurin series, that is, 9ex = [infinity] 9xn n! n = 0 for all x.