Find a limit $ \lim_{c\to\infty} E\big[ X e^{-\frac{(X-c)^2}{2}} \big] \big/ E\big[e^{-\frac{(X-c)^2}{2}}\big]$ Let $X$ be a bounded random variable, i.e., $|X| \le a$. 
Can we find
\begin{align}
g(c)=\frac{E\left[  X e^{-\frac{(X-c)^2}{2}} \right]}{E\left[e^{-\frac{(X-c)^2}{2}}\right]}
\end{align}
How to find the following limit
\begin{align}
\lim_{ c \to \infty} g(c)
\end{align}
I tried L'Hopital rule but it did not work out. 
 A: The intuition comes from the special case where $X$ is discrete and takes only finitely many values with probability 1. Indeed, assume that $x_1 < \cdots x_n$ and $p_i := \Bbb{P}(X = x_i) > 0$ for each $i = 1, \cdots, n$ and $p_1 + \cdots + p_n = 1$. Then
$$ g(c)
= \frac{\sum_{i=1}^{n} x_i p_i e^{-(c-x_i)^2/2}}{\sum_{i=1}^{n} p_i e^{-(c-x_i)^2/2}} 
\xrightarrow[\ c\to\infty \ ]{} x_n = \operatorname{esssup}X.$$
For general cases, let $\alpha < \beta < \operatorname{esssup}X$ and consider
$$ M(c) = \Bbb{E}[e^{-(X-c)^2/2} \mathbf{1}_{\{X > \alpha\}} ]
\quad \text{and} \quad
m(c) = \Bbb{E}[e^{-(X-c)^2/2} \mathbf{1}_{\{X \leq \alpha\}} ]. $$
We assume $c > a$ WLOG so that $x \mapsto e^{-(x-c)^2/2}$ is increasing on $[-a, a]$. Then from the following simple estimates
$$
M(c)
\geq \Bbb{E}[e^{-(X-c)^2/2} \mathbf{1}_{\{X \geq \beta\}} ]
\geq e^{-(\beta-c)^2/2} \Bbb{P}(X \geq \beta)
\quad \text{and} \quad
m(c) \leq e^{-(\alpha-c)^2/2}
$$
we easily find that
$$ \frac{m(c)}{M(c)} \leq \frac{e^{(\beta^2-\alpha^2)/2}}{\Bbb{P}(X \geq \beta)} e^{-c(\beta-\alpha)} \xrightarrow[ \ c \to \infty \ ]{} 0. $$
Now by writing
$$ g(c)
= \frac{\Bbb{E}[X e^{-(X-c)^2/2} \mathbf{1}_{\{X > \alpha\}}] + \Bbb{E}[X e^{-(X-c)^2/2} \mathbf{1}_{\{X \leq \alpha\}}]}{\Bbb{E}[e^{-(X-c)^2/2} \mathbf{1}_{\{X > \alpha\}}] + \Bbb{E}[e^{-(X-c)^2/2} \mathbf{1}_{\{X \leq \alpha\}}]}
\geq \frac{\alpha M(c) - a m(c)}{M(c) + m(c)} $$
it follows that
$$ \alpha \leq \liminf_{c\to\infty} g(c) \leq \limsup_{c\to\infty} g(c) \leq \operatorname{esssup} X. $$
Taking $\alpha \uparrow \operatorname{esssup}X$ completes the proof.
