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.