This was a typo. Sub $\displaystyle y=\frac{\pi}{L}x$ and $\displaystyle dx=\frac{\pi}{L}dy$, in order to change the limits of integration from $\displaystyle [-L,L]$ to $\displaystyle [-\pi,\pi]$ in accordance with the R-L lemma. Thanks for catching that blooper, Sampras.

Yes it does. Completing the RHS of the substitution, $\displaystyle

\frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy=f(0)

$

Regardless of what f is, it needs to sit on the origin. Here is a proof: If $\displaystyle \int_{-a}^a h(x)dx=0$ for all $\displaystyle a\in\mathbb{R}$, then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that $\displaystyle h(-x)=h(x)$ but $\displaystyle \lim_{n\rightarrow\infty} h(x)-h(-x)$ does need to be zero.

Consider this counterexample: $\displaystyle f(x)=x+1$, so $\displaystyle f(0)\neq0$ Theorem: $\displaystyle \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l \frac{x+1}x sin(nx) dx\neq0$

Proof: $\displaystyle \lim_{n\rightarrow\infty} \frac{x+1}x sin(nx) - \frac{-x+1}{-x} sin(-nx)$=$\displaystyle \lim_{n\rightarrow\infty} \frac{\sin(nx)}x [(x+1)-(1-x)]$=$\displaystyle \lim_{n\rightarrow\infty} 2\sin(nx)\neq 0$

*I'm not sure how rigorous this is, but because the $\displaystyle \frac{x+1}x sin(nx)$ is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.