Convergence of fourier series of a non continuous function Question:
Let $g: [-\pi,\pi] \to \Bbb R$ is defined:
$\begin{cases} 5+\tan \frac 1{3x} & x\ne0\\ 5&x=0 \end{cases}$
let $f \in \Bbb R (\Bbb T)$ be the periodic continuation of $g$ from $[-\pi,\pi)$ to $\Bbb R$. Prove that the Fourier series of f covnerges in $x=0$ to $f(0)$.
Thoughts
We cannot use Dirichlet because the function is not continuous and the one sided 
limits don't exist. 
The Fourier series itself is pretty hard to compute- is there some trick involved?
Would love some hints first. 
 A: Igor Rivin pointed out a  problem with this question: the Fourier series (in the usual sense) is not even defined for $g$. But we can try to make some sense of it by  taking divergent integrals in the sense of principal value (with infinitely many p.v. points!?), or better, by truncating $\tan \frac{1}{3x}$. Truncation means we  replace $\tan\frac{1}{3x}$ with
$$\max\left(-M, \min\left(M, \tan \frac{1}{3x}\right)\right)$$
compute the integrals and then let  $M\to \infty$. The key points are: 


*

*$5$, being constant, contributes only to the constant term of the Fourier series 

*$\tan(1/3x)$, being odd (also after truncation), contributes only to the sine terms of the Fourier series.


Thus, the series has no terms $\cos nx$ with $n\ne 0$.
All  sine terms all equal to $0$ when $x=0$. Which leaves you with $5$. 
A: Well, you can make a sequence $g_n(x),$ where $g_n$ agrees with $g$ for $|x|>\frac{1}{n\pi},$ and is equal to $5$ otherwise. The fourier series of that can be (in principle) computed, and by Dirichlet equals to $g_n$ near the origin. Then use dominated convergence on $g_n$ or fourier coeff's thereof?!
edit Actually, there is a problem: $\tan \frac{1}{3 x}$ is not (except in a principal value way) integrable on any interval containing $0.$ So, the question is, what Fourier series? 
