Let $X$ be a real-valued random variable with $EX=0$, $VarX=\sigma^2<\infty$. Prove $$\phi_X(t)=1-\frac{t^2\sigma^2}{2}+\mathcal{o}(t^2) \text{ as } t\to0$$ with $\phi_X$ the characteristic function.
We have a theorem: if $E|X|^n<\infty$ then $$\phi_X(t)=\sum_{r=0}^n\frac{(it)^r}{r!}EX^r+\frac{(it)^r}{r!}\varepsilon_n(t)$$ with $\varepsilon_n(t)\to0$ as $t\to0$. For $n=2$ it is $$\phi_X(t)=EX^0-\frac{t^2\sigma^2}{2}+\frac{(it)^2}{2}\varepsilon_2(t)$$ since $EX^1=0, EX^2=\sigma^2$. But what is $EX^0$ and how do I prove $\mathcal{0}(t^2)$? Thanks for any help!
First note that, since $E(X)=0$, $\mathrm{var}(X) = E(X^2) -E(X)^2 = E(X^2)$.
Now you can use your theorem with $n=2$ as you are doing.
To obtain the final assertion, first note that $E(X^0)=E(1)=1$ by normalization. We are left with showing that the error is $o(t^2)$.
The definition of little-$o$ is the following. We say $f(t)=o(g(t))$ for $t\to t_0$ if
$$ \lim_{t\to t_0} \frac{f(t)}{g(t)} = 0. $$
You have a little typo in your theorem, the correct expansion is
$$ \phi_X(t)=\sum_{r=0}^n\frac{(it)^r}{r!}E(X^r)+\frac{(it)^n}{n!}\varepsilon_n(t). $$
If you use this expansion with $n=2$ the error term is given by
$$ \mathrm{Err}(t) = -\frac{t^2}{2} \varepsilon_2(t)$$
And since
$$ \lim_{t\to 0} \frac{\mathrm{Err}(t)}{t^2} = \lim_{t\to 0} -\frac{1}{2}\varepsilon_2(t) = 0 $$
you see that $\mathrm{Err}(t) = o(t^2)$.