Final answer:
The sequence (f_n)_n, defined by f_n(x) = cos(nx)/n, converges uniformly to the zero function as n approaches infinity because the amplitude of the cosine function decreases to zero for all x.
Step-by-step explanation:
To show that the sequence of functions (f_n)_n, defined by f_n(x) = cos(nx)/n, converges uniformly to a function f, we need to prove that for all ε > 0, there exists an N such that for all n ≥ N and all x in the domain of f_n, the absolute value of f_n(x) - f(x) is less than ε. Since we have a case of a trigonometric function and its amplitude being scaled down by n, we know that the cosine function has a bounded range of [-1, 1]. Therefore, |cos(nx)| ≤ 1 for all x.
As n increases, 1/n approaches zero, so no matter the value of x, |f_n(x)| = |cos(nx)/n| ≤ 1/n becomes arbitrarily small. Hence, for any given ε > 0, we can choose N such that 1/N < ε and then for all n ≥ N, |f_n(x) - f(x)| = |f_n(x)| since f(x) = 0 < ε. This implies the sequence converges uniformly to the zero function f(x) = 0.