= Solution
Finite <second moment> implies $\mathbb EX<\infty$. For $\lambda$ near any positive $\lambda_0$, the <derivative> of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty)$. The <dominated convergence theorem> therefore permits <differentiation> under the <expectation>:
$$
\boxed{\phi_X'(\lambda)=-\mathbb E[Xe^{-\lambda X}],\qquad\lambda>0.}
$$
For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. <Taylor expansion> gives $r(z)/z^2\to0$ as $z\downarrow0$. Also, for $z\geq0$,
$$
0\leq e^{-z}-1+z\leq\frac{z^2}{2},
\qquad |r(z)|\leq\frac{z^2}{2}.
$$
The inequalities follow by integrating $0\leq1-e^{-u}\leq u$ from zero to $z$. Consequently $r(\lambda X)/\lambda^2\to0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the <dominated convergence theorem> gives the <second-moment expansion of a Laplace transform>:
$$
\boxed{\phi_X(\lambda)=1-\lambda\mathbb EX
+\frac{\lambda^2}{2}\mathbb EX^2+o(\lambda^2),\qquad\lambda\downarrow0.}
$$
No third moment is needed for the remainder estimate.
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