Final answer:
To prove uniform convergence of the sequence ((nx² - x - 1/n) e^{-nx}) on [0, ∞), we find that its pointwise limit is 0. Demonstrating that the absolute difference |(nx² - x - 1/n) e^{-nx}| reaches a maximum depending on n and approaches 0 as n increases, we show uniform convergence.
Step-by-step explanation:
To prove that the sequence ((nx² - x - 1/n) e^{-nx}) converges uniformly on [0, ∞), we need to show that for every ε > 0, there exists an N such that for all n ≥ N and all x in [0, ∞), the absolute difference between f_n(x) (the nth term of the sequence) and the limit function f(x) is less than ε. First, we determine the pointwise limit of the sequence as n tends to infinity. For all x ≥ 0, as n → ∞, e^{-nx} → 0. Thus, the pointwise limit of f_n(x) is 0. Next, to show uniform convergence, consider the absolute difference |f_n(x) - f(x)| = |(nx² - x - 1/n) e^{-nx}|. Since e^{-nx} is always positive, we can ignore the exponential part for the maximum value. The quadratic term nx² - x - 1/n achieves its maximum value when its derivative with respect to x, 2nx - 1, equals zero, which gives x = 1/(2n). Substituting back, we get a maximum difference that depends solely on n, and as n increases, this difference approaches 0. Hence, f_n(x) converges uniformly to f(x).