Verified answer

Let's try the Mean Value Theorem (assuming n > 0).

f(x) - f(y) = f '(c) (x - y) for some c between x and y.

.............= [-n^2 cos^(n-1)(c) sin^2(c) + n cos^(n+1)(c)] (x - y)

.............= [-n^2 cos^(n-1)(c) (1 - cos^2(c) + n cos^(n+1)(c)] (x - y)

.............= [-n^2 cos^(n-1)(c) + (n^2 + n) cos^(n+1)(c)] (x - y)

Hence,

|f(x) - f(y)| ≤ [n^2 |cos^(n-1)(c)| + (n^2 + n) |cos^(n+1)(c)|] |x - y|, by triangle inequality

...............≤ [n^2 * 1 + (n^2 + n) * 1] |x - y|

...............= (2n^2 + n) |x - y|.

So given ε > 0, let δ = ε/(2n^2 + n). Then for all x, y with |x - y| < δ, we have

|f(x) - f(y)| ≤ (2n^2 + n) |x - y| < (2n^2 + n) * ε/(2n^2 + n) = ε, as required.

I hope this helps!