Is $\lim \limits_{n\to\infty} \int_0^1 \sin(\frac{1}{x}) \sin(nx)dx$ convergent and if so, what is the limit?
Neither Riemann-Lebesgue lemma nor Dirichlet lemma can be applied directly. The limit seems to be 0, but I'm not completely certain. Dirichlet lemma is as follows.
Let $f:(0,1)\to \mathbb{R}$ be monotone and bounded. Then $\lim \limits_{n\to\infty}\int_0^1f(t)\frac{sin(tn)}{t}dt=\frac{\pi}{2}\lim \limits_{t\to 0^+}f(t).$
On
Here a quantitative and explicit approach.
$$\int_{0}^{1}\cos\left(\frac{1}{x}-nx\right)\,dx\stackrel{\frac{1}{x}-nx\mapsto z}{=}\frac{1}{2n}\int_{n-1}^{+\infty}\cos(z)\left(1-\frac{z}{\sqrt{z^2+4n}}\right)\,dz $$ equals $$ 2\int_{n-1}^{+\infty}\frac{\cos(z)\,dz}{\sqrt{z^2+4n}(z+\sqrt{z^2+4n})}=\\=2\int_{n-1}^{+\infty}\frac{dz}{\sqrt{z^2+4n}(z+\sqrt{z^2+4n})} -2\int_{n-1}^{+\infty}\frac{(1-\cos(z))\,dz}{\sqrt{z^2+4n}(z+\sqrt{z^2+4n})}$$
where the last integral is clearly positive but also bounded by $4\int_{n-1}^{+\infty}\frac{dz}{\sqrt{z^2+4n}(z+\sqrt{z^2+4n})}=\frac{2}{n}$, so $$ \left|\int_{0}^{1}\cos\left(\frac{1}{x}-nx\right)\,dx\right|\leq \frac{1}{n}. $$ Dealing with $\int_{0}^{1}\cos\left(\frac{1}{x}+nx\right)\,dx$ is a bit more involved since $\frac{1}{x}+nx$ has a stationary point at $x=\frac{1}{\sqrt{n}}$. Anyway
$$ \int_{0}^{1/\sqrt{n}}\cos\left(\frac{1}{x}+nx\right) = -\frac{1}{2n}\int_{2\sqrt{n}}^{+\infty}\cos(z)\left(1-\frac{z}{\sqrt{z^2-4n}}\right)\,dz $$ by the same approach turns to be bounded (in absolute value) by $\frac{1}{\sqrt{n}}$, and the same applies to $$ \int_{1/\sqrt{n}}^{1}\cos\left(\frac{1}{x}+nx\right)\,dx = -\frac{1}{2n}\int_{2\sqrt{n}}^{n+1} \cos(z)\left(1-\frac{z}{\sqrt{z^2-4n}}\right)\,dz, $$
so for any $n\geq 2$
$$ \left|\int_{0}^{1}\sin\left(\frac{1}{x}\right)\sin(nx)\,dx\right|\leq \frac{1}{\sqrt{n}}+\frac{1}{2n}. $$
$\int_0^{1}|\sin (\frac 1 x)| dx=\int_1^{\infty} \frac { |\sin y|} {y^{2}}dy <\infty$. So $\sin (\frac 1 x)$ is integrable on $(0,1)$ and Riemann Lebesgue Lemma shows that the limit is $0$.