Does the Fourier series converge at $x=0$? 
Let $f(x)$, a $2\pi$ periodic funciton such that $f(0) = 1$ and for every $0\ne x\in[-\pi,\pi]$: $f(x) = 1 + \sin \frac{\pi^2}{x}$. Is the Fourier series of $f(x)$ converges at $x=0$? If so, what it's value there?

Well, $\lim_{x\to 0} f(x)$ doesn't exist. We have learned in class that if a function is continuous at a point $x_0$ (or even locally Lipchitz) then it's Fourier series converges to $f(x)$. Moreover, we have learned that if the sided limits exist but different then the Fourier series converges to $$\frac{f(x^+)+f(x^-)}{2}.$$
But here, the limit doesn't even exist. So is it implying the Fourier series doesn't converge to $f(0)$?
 A: 
So is it implying the Fourier series doesn't converge to $f(0)$?

No, it doesn't imply that, and in this case, the Fourier series of $f$ does converge to $f$ pointwise.
If we look at $g(x) = f(x) - 1$, we note that $g$ is an odd function, hence in the real Fourier series, only sine terms occur, and of course $\sin (nx) = 0$ for $x = 0$, whence the Fourier series of $g$ converges to $g$ at $0$. For $0 < \lvert x\rvert < \pi$, the continuous differentiability of $g$ in a neighbourhood of $x$ implies the (pointwise) convergence of the Fourier series, and at $\pm \pi$, we have $g(\pm\pi) = 0$, and $\sin (\pm n\pi) = 0$ for all $n$, so pointwise convergence there too (that also follows from $g$ being locally Lipschitz in a neighbourhood of $\pm\pi$).
Adding the constant term $\cos (0x)$ to get the Fourier series of $f$ doesn't change the convergence behaviour.
A: Examine pointwise convergence condition for a Fourier series.  Note that ypur function though not continuous at 0 is Lebesgue integrable in $[- \pi, \pi]$.
Let me state the point wise convergence criterion.
Theorem.  Suppose $f$ is Lebesgue integrable in $[-\pi, \pi]$.    A necessary and sufficient condition for the Fourier series $T(x)$ of  $f$  to converge pointwise to $c$ at $x$ is that there exists a fixed $\delta$ such that $0 < \delta < \pi $ and
$\int_0^\delta  {{g_c}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ pointwise as $n$ tends to ∞ , where   ${g_c}(u) = \frac{1}{2}\left( {f(x + u) + f(x - u) - 2c} \right)$ .
For a reference see Theorem 19 (i) in my article Convergence of Fourier series

From this theorem you can deduce all the usual statement about convergence when $f$ is continuous at $x$ or when $f$ has a jump discontinuity at $x$.
One simple deduction is when $f$ is a constant c + an odd function, for $x = 0$,
the Fourier series at $x = 0$ converges to c.  This is because for $x = 0$ ,
${g_c}(u) = \frac{1}{2}\left( {f(u) + f( - u) - 2c} \right) = \frac{1}{2}\left( {c + f(u) + c - f(u) - 2c} \right) = 0$ . 
We can take any $0 < \delta < \pi$ .
Consequently and trivially, $\int_0^\delta  {{g_c}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ pointwise as $n$ tends to ∞ . 
Your function $f(x) = 1 + \sin \left( {{\textstyle{{{\pi ^2}} \over x}}} \right)$ is $1$ + an odd function and so its Fourier series at $x = 0$ converges pointwise to the value 1.
At all other points other than 0 in $[-\pi, \pi]$, $f$ is differentiable and so the Fourier series of $f$ at $x$ for $x$ in $[-\pi, \pi] - {0}$ converges poinwise to $f(x)$.  This is because the limit
$\mathop {\lim }\limits_{u \to 0} \frac{{f(u + x) - f(x)}}{u} = f'(x)$ exists
and as a consequence
$\frac{{{g_{f(x)}}(u)}}{u} = \frac{1}{2}\left( {\frac{{f(x + u) + f(x - u) - 2f(x)}}{u}} \right)$ is Lebesgue integrable.
 Therefore, by the Riemann Lebesgue Lemma,
 $\int_0^\pi  {{g_{f(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ .  The pointwise convergence of the Fourier series at $x$ not equal to $0$ then follows from the above theorem.
(Note that you don't need continuous differentiability at $x$ for $g$ to be Lebesgue integrable.)  
