Log of the Moment Generating Function

Question

The Question:

Let $X$ be a random variable such that $\Bbb E[X] = 0$ and $\text{Var}(X)=1$. Show that, as $t \rightarrow 0$,

$$\ln M_X(t)= c t^2 + o(t^2)$$

where $M_X$ is the MGF of $X$, $o$ is the "small-oh" notation, and $c$ is a constant you should determine.

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My Attempt:

So I begun by Taylor expanding $M_X(t)$ about $t=0$, which gives

\begin{align} M_X(t) & = M_X(0)+M'_X(0)t+M''_X(0) \frac{t^2}{2}+O(t^3) \\ & = 1+\Bbb E[X]t+\Bbb E[X^2]\frac{t^2}{2} + O(t^3) \\ & = 1+(0)t+(1+0^2)\frac{t^2}{2}+O(t^3) \\ & = 1+\frac{t^2}{2}+O(t^3) \end{align}

and then how do I take natural logs, let $t \rightarrow 0$ and still have $t$ in the expression?

Answer 1

Note $\ln(1+x)=x+O(x^2)$, so $$ \ln M_X(t)=\ln(1+t^2/2+o(t^2))=(t^2/2+o(t^2))+O((t^2/2+o(t^2))^2) $$ Can you take it from here?