Correct answer: (A) S is equicontinuous.; (C) S is bounded.
Explanation
For t=nx≥0, |f_n′|=|1−t|e⁻ᵗ≤1, so the family is uniformly Lipschitz, proving A. Supremum norm is 1/(en)≤1/e, proving C. These norms tend to zero, but the zero function is not in S. Thus S is not closed and cannot be compact in this metric space. **A,C**.
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