I lost 18 out of 20 points on this question and I'm not sure why.
Suppose $f$ is continuous on $\left[0,\frac{\pi}{2}\right]$. What is the $\lim_{n \to \infty} \int\limits_0^\frac{\pi}{2} f(x)\sin(nx)\,dx$? Prove your claim. Hint: Use Stone-Weierstrass
Here is my proof:
Let $\epsilon > 0$ and consider $\mathcal{A}= \left\{\sum_{n=1}^Na_n\sin(nx)\,|\,n\in\mathbb{N}, a_n\in\mathbb{R}\right\}$ on $\left[\epsilon,\frac{\pi}{2}\right]$.
$\mathcal{A}$ is an algebra of functions which vanishes only at $x=0$, and which separates points, since $\sin(x)$ is a 1-1 function on $\left[\epsilon,\frac{\pi}{2}\right]$. Therefore, by the Stone-Weierstrass Theorem, there is a sequence $s_n$ of functions of $\mathcal{A}$ that converges uniformly to $f(x)$ on $\left[\epsilon,\frac{\pi}{2}\right]$ for every $\epsilon>0$.
Since $s_n$ is a series that converges uniformly to $f(x)$, there is a $N\in\mathbb{N}$ such that when $N<m<n$ we have $|\sum_{m=i}^na_i\sin(ix)|<\epsilon|a_n|$, and when $n=m+1$, $|a_n\sin(nx)|\leq\epsilon |a_n|$ for $x\in\left[\epsilon,\frac{\pi}{2}\right]$
Therefore
$$|\sin(nx)|\leq\epsilon$$
$$|f(x)\sin(nx)|\leq|f(x)|\epsilon=|f(x)\epsilon|$$
$$\int\limits_\epsilon^\frac{\pi}{2} |f(x)\sin(nx)|\,dx\leq \int\limits_\epsilon^\frac{\pi}{2} |f(x)\epsilon|\,dx = \epsilon\left[F\left(\frac{\pi}{2}\right)-F(\epsilon)\right],$$
where $F'(x)=f(x)$.
By the Fundamental Theorem of Calculus, $\int\limits_x^\frac{\pi}{2} f(t)\sin(nt)\,dt$ is continuous, so for sufficiently large $n$,
$$\lim_{\epsilon \to 0} \int_\epsilon^\frac{\pi}{2} |f(x)\sin(nx)|\,dx=\lim_{\epsilon \to 0}\epsilon\left[F\left(\frac{\pi}{2}\right)-F(0)\right]= 0\left[F\left(\frac{\pi}{2}\right)-F(0)\right]=0.$$
So $\lim_{n \to \infty} \int_0^\frac{\pi}{2} f(x)\sin(nx)\,dx=0$.