If $$\lim_{c \to 0} \frac{\phi_X(c) - 1}{c^2} = -\frac{\sigma^2}{2} < \infty$$ where $\phi_X(c)$ is the characteristic function of the random variable $X$, then $$E[X] = 0,\qquad E[X^2] = \sigma^2$$
I know $$\lim_{c \to 0} \frac{\phi_X(c) - 1}{c^2} = \lim_{c \to 0} \frac{E[e^{icX}] - 1}{c^2}$$ Also, I've proven that if the $n$-th moment exists, the $n$-th derivative of the characteristic function exists, and this seems a bit like a partial converse, but I can't come up with the right angle to look at the limit from to see how the suggested values for $E[X]$ and $E[X^2]$ come out of it. I would like just a hint on how I should be looking at this problem.
First, we have to show that the second moment is finite. Notice that for any $x$, $2n^2(1-\cos(x/n))\to x^2$, hence $$\mathbb E[X^2]=\mathbb E\left[\liminf_{n\to \infty}2n^2(1-\cos(X/n))\right].$$As a consequence of Fatou's lemma, we obtain that $$\mathbb E[X^2]\leqslant \liminf_{n\to \infty}2n^2\mathbb E\left[(1-\cos(X/n))\right].$$ Since $$2\mathbb E\left[(1-\cos(X/n))\right]=2-\phi(1/n)-\phi(-1/n),$$ we get $$\liminf_{n\to \infty}2n^2\mathbb E\left[(1-\cos(X/n))\right]\leqslant \sigma^2.$$ We thus are allowed to compute the expectation taking the derivative of the characteristic function. Using the assumption, the derivative is $0$. Since $\phi''(0)=-\mathbb E[X^2]$, then we get $\mathbb E[X^2]=\sigma^2$.
We can also deduce an asymptotic expansion at $0$, see here.