Limit Relation Involving Second Moment Saw this limit relation in a paper, but I can't seem to prove it:
\begin{equation}
\mathbb{E}(X^2) = \lim_{t \rightarrow 0} \frac{2}{t^2} [\mathbb{E}_X(e^{tX})-1]
\end{equation}
I might be missing something basic, but does anyone know how? Not sure if we need to assume $\mathbb{E}(X)=0$ here too.

$$
e^{tX} = 1 + tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
$$
e^{tX} - 1 = tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots
$$
\begin{align}
\operatorname{E}(e^{tX} - 1) & = \operatorname{E}\left( tX + \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots \right) \\[10pt]
& = \operatorname{E} \left( \frac{t^2 X^2} 2 + \frac{t^3 X^3} 6 + \frac {t^4 X^4}{24} + \frac{t^5 X^5}{120} + \cdots \right) \text{ if } \operatorname{E}(X)=0 \\[10pt]
\frac 2 {t^2} \operatorname{E}(e^{tX} - 1) & = \operatorname{E} \left( X^2 + \frac{t X^3} 3 + \frac {t^2 X^4}{12} + \frac{t^3 X^5}{60} + \cdots \right) \text{ if } \operatorname{E}(X) = 0.
\end{align}
I think L'Hopital's rule could also be used.
