How do I Evaluate $\lim_{n \rightarrow \infty} \int_{[0,1]} x\sin(\frac{n}{x})dx$? Would a viable strategy be to subtitute $y=\frac{n}{x}$ and then use integration by parts? I'm not sure If i'm able to do this, what are the technical difficulties that I need to be aware of? 
I'm so poor at Lebesgue integration >.<. I would really appreciate somebody taking the time to wal me through this.. :-(
 A: Your original idea works fine.  If you are concerned about the (removable) singularity of the integrand , then simply write
$$\begin{align}
\require{cancel}
\lim_{n\to\infty}\lim_{\varepsilon\to 0}\int_\varepsilon^1 x\sin(n/x)\,dx&=\lim_{n\to\infty}\lim_{\varepsilon\to 0}\int_n^{n/\varepsilon} \frac{\sin(x)}{x}\,dx\\\\
&=\lim_{n\to\infty}\lim_{\varepsilon\to 0}\left.\left(\frac{-\cos(x)}{x}\right)\right|_{x=n}^{x=n/\varepsilon}-\lim_{n\to\infty}\lim_{\varepsilon\to0}\int_n^{n/\varepsilon}\frac{\cos(x)}{x^2}\,dx\\\\
&=\lim_{n\to\infty}\lim_{\varepsilon\to 0}\cancelto{0}{\left(\frac{\cos(n)}{n}-\frac{\varepsilon\cos(n/\varepsilon)}{n}\right)}-\lim_{n\to\infty}\lim_{\varepsilon\to0}\int_n^{n/\varepsilon}\frac{\cos(x)}{x^2}\,dx\\\\
&=-\lim_{n\to\infty}\lim_{\varepsilon\to0}\int_n^{n/\varepsilon}\frac{\cos(x)}{x^2}\,dx\\\\
\end{align}$$
Noting that $\left|\int_n^{n/\varepsilon}\frac{\cos(x)}{x^2}\,dx\right|\le \frac1n-\frac{\varepsilon}{n}$, we find that the coveted limit of the integral of interest is $0$.  And we are done.
