Let $X\sim \mu$ be a random variable taking value on $R$. Assume that $d\mu=C\sqrt{4-x^2}dx $ for some normalized constant $C>0$. Assume that the limit of moment generating function $$ \lim_{t\to \infty}\frac{E[e^{tX}]}{t^{-3/2}e^{2t}}=1. $$
Can we use the above result to get the asymptotic limit of the derivative of the moment generating function $\frac{d}{dt} E[e^{tX}]=E[Xe^{tX}]$?
I try to find a relation between $E[Xe^{tX}]$ and $E[e^{tX}]$.
Note that $$ E[e^{tX}]=C\int_{-2}^2e^{tx}\sqrt{4-x^2}dx. $$
After giving it a try I found that it's probably easier to deal with hypergeometric functions (which encompass all Bessel functions): start by representing $X$ as $2(2Y-1)$ where $Y$ has a Beta$(\frac32,\frac32)$ distribution. Then $$M(t):=E\!\left[\mathrm e^{tX}\right]=\mathrm e^{-2t}E\!\left[\mathrm e^{4tY}\right]=\mathrm e^{-2t}\cdot{}_1F_1(\tfrac32;3;4t),$$ where $_1F_1$ is Kummer's confluent hypergeometric function. Its asymptotic behavior is known: $$_1F_1(\tfrac32;3;4t)\sim\frac{\Gamma(3)}{\Gamma(\tfrac32)}\cdot\mathrm e^{4t}(4t)^{-\frac32}=\frac{\mathrm e^{4t}t^{-\frac32}}{2\sqrt\pi}.\tag{1}$$ Therefore $$\lim_{t\to\infty}\frac{M(t)}{t^{-\frac32}\mathrm e^{2t}}=\color{red}{\frac1{2\sqrt\pi}}$$ (and not $1$). For the derivative, we have $$M'(t)=\mathrm e^{-2t}\left(-2\cdot{}_1F_1(\tfrac32;3;4t)+4\cdot\partial_t\,{}_1F_1(\tfrac32;3;4t)\right).$$ The derivative of Kummer's function gives $$ \partial_t\,{}_1F_1(\tfrac32;3;4t)=-\frac12\cdot{}_1F_1(\tfrac32;4;4t)+{}_1F_1(\tfrac32;3;4t), $$ where ${}_1F_1(\tfrac32;4;4t)=O\bigl(\mathrm e^{4t}t^{-\frac52}\bigr)$ is negligible as $t\to\infty$ compared to (1). Then $$ \partial_t\,{}_1F_1(\tfrac32;3;4t)\sim{}_1F_1(\tfrac32;3;4t).$$ Thus we can conclude that $M'(t)\sim 2M(t)$, and $$\lim_{t\to\infty}\frac{E[X\mathrm e^{tX}]}{t^{-3/2}\,\mathrm e^{2t}}=\frac1{\sqrt\pi}.$$