Uniform Convergence of $\sin^2 (x +1/n )$

Question

For $n\geq 1$ and let $g_n(x) = \sin^2(x+ \frac{1}{n}) , x\in [0,\infty)$ and $f_n(x)=\int_{0}^{x}g_n(t)dt$ , then

1. {$f_n(x)$} converges point wise to a function on $[0,\infty)$ , but does not converge uniformly on $[0,\infty)$
2. {$f_n(x)$} does not converge point wise to any function on $[0,\infty)$
3. {$f_n(x)$} converges uniformly to a function on $[0,1]$
4. {$f_n(x)$} converges uniformly to a function on $[0,\infty)$

What I answered is {$f_n(x)$} converges uniformly to a function on $[0,\infty)$ because the limit converges to $\sin^2(x)$ which is bounded continuous function

Answer 1

Right answer, wrong reason. Yes, the limit function is bounded and continuous, but that does not imply the convergence is uniform. For example, $f_n(x) = x/n$ converges pointwise to $0$ on $[0,\infty),$ but the convergence is not uniform there ($\sup_{[0,\infty)} |f_n| = \infty$ for all $n$).

Hint to see that $\int_0^x \sin^2(t+1/n)\,dt$ converges uniformly to $\int_0^x \sin^2(t)\,dt$ on $[0,\infty):$

$$\int_0^x \sin^2(t+1/n)\,dt = \int_{1/n}^{x+1/n} \sin^2(t)\,dt.$$