Show $ \sqrt{p}Y_{p}\overset{d}{\rightarrow} L(a)$ as $p\rightarrow 0$ 
Suppose that $X_{1}, X_{2}, \ldots $ are i.i.d. symmetric random variables with the finite variance $\sigma^{2}$, let $N_{p}\in \text{Fs}(p)$ be independent of $X_{1},X_{2},\ldots$, and set $Y_{p}=\sum_{k=1}^{N_{p}}X_{k}$. Show that
\begin{equation*}
\begin{split}
\sqrt{p}Y_{p}\overset{d}{\rightarrow} L(a)\qquad \text{as}\quad p\rightarrow 0.
 \end{split}
  \end{equation*}
and determine $a$.

Attempt at solution:
$ \psi_{X}(t)=\mathrm{E}[e^{tX}]\\
\psi_{X}(\sqrt{(0)}t))=1\\
\frac{d}{dp}\psi_{X}(\sqrt{p}t))=\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)\\
g_{N_{p}}(t)=\sum_{k=1}^{\infty}t^{k}(1-p)^{k-1}p=\frac{p}{1-p}\sum_{k=1}^{\infty}t^{k}(1-p)^{k}\\
=\frac{p}{1-p}\sum_{k=1}^{\infty}(t(1-p))^{k}=\frac{p}{1-p}(\frac{1}{1-(t(1-p)}-\frac{1-(t(1-p)}{1-(t(1-p)})\\
=\frac{p}{1-p}(\frac{(t(1-p)}{1-(t(1-p)})=\frac{pt}{1-(1-p)t}\\
\sqrt{p}Y_{p}= \sqrt{p}\sum_{k=1}^{N_{p}}X_{k}=\sum_{k=1}^{N_{p}}(\sqrt{p}X_{k})\\
\psi_{\sqrt{p}Y_{p}}(t)=g_{N_{p}}(\psi_{X}(\sqrt{p}t))=\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\
\lim_{p\rightarrow 0}\sqrt{p}Y_{p}=\lim_{p\rightarrow 0}\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\
\text{indeterminate, use l'hopital}\qquad \frac{0}{0}\\
=\lim_{p\rightarrow 0}\frac{\frac{d}{dp}}{\frac{d}{dp}}\frac{p\psi_{X}(\sqrt{p}t)}{1-(1-p)\psi_{X}(\sqrt{p}t)}\\
  =\lim_{p\rightarrow 0}\frac{p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}{-\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}\\
   =\lim_{p\rightarrow 0}\frac{p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t)}{\frac{1}{2\sqrt{p}}(-t\psi_{X}'(\sqrt{p}t)+pt\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t))}\\
=\lim_{p\rightarrow 0}\frac{2\sqrt{p}(p\frac{t}{2\sqrt{p}}\psi_{X}'(\sqrt{p}t)+\psi_{X}(\sqrt{p}t))}{(p-1)t\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}\\
=\lim_{p\rightarrow 0}\frac{tp\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}{(p-1)t\psi_{X}'(\sqrt{p}t)+2\sqrt{p}\psi_{X}(\sqrt{p}t)}\\
$
At this point I'm stuck, this limit goes to $\frac{0}{t\psi_{X}'(\sqrt{p}t)}=0$. I guess that I am supposed to use the fact that the distribution of $X_{k}$ is symmetric, but cannot figure out how.
 A: I worked on it some more and got this solution:
\begin{equation*}
    \begin{split}
\varphi_{X}(t)=\mathrm{E}(e^{itX})\\
 Y_{p}=\sum_{k=1}^{N_{p}}X_{k}=(\varphi_{X_{1}}(t))^{N_{p}}
     \end{split}
      \end{equation*}
Since
\begin{equation*}
\begin{split}
\varphi_{X}(t)=1+\sum_{k=1}^{n}X^{k}\cdot \frac{(it)^{k}}{k!}+\mathscr{o}(|t|^{n})\quad \text{as}\quad t\rightarrow 0
 \end{split}
  \end{equation*}
we get that
\begin{equation*}
\begin{split}
 \varphi_{X}(\sqrt{p} t)=1+\sum_{k=1}^{n}X^{k}\cdot \frac{(i\sqrt{p}t)^{k}}{k!}+\mathscr{o}(|\sqrt{p}t|^{n})\quad \text{as}\quad p\rightarrow 0
 \end{split}
  \end{equation*}
Thereby
\begin{equation*}
\begin{split}
\sqrt{p}Y_{p}=\sum_{k=1}^{N_{p}}\sqrt{p}X_{k}=(\varphi_{X_{1}}(\sqrt{p}t))^{N_{p}}=(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\mathrm{E}X^{2}+\mathscr{o}(pt^{2}))^{N_{p}}\\
=(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))^{N_{p}}
\end{split}
  \end{equation*}
\begin{equation*}
\begin{split}
 g_{N_{p}}(t)=\mathrm{E}(t^{N_{P}})=\sum_{k=1}^{\infty}t^{k}(1-p)^{k-1}p=\frac{p}{1-p}\sum_{k=1}^{\infty}t^{k}(1-p)^{k}\\
=\frac{p}{1-p}\sum_{k=1}^{\infty}(t(1-p))^{k}=\frac{p}{1-p}(\frac{1}{1-(t(1-p)}-\frac{1-(t(1-p)}{1-(t(1-p)})\\
=\frac{p}{1-p}(\frac{(t(1-p)}{1-(t(1-p)})=\frac{pt}{1-(1-p)t}\\
\end{split}
  \end{equation*}
\begin{equation*}
\begin{split}
\varphi_{\sqrt{p}Y_{p}}(t)=g_{N_{p}}( \varphi_{X}(\sqrt{p} t))=\mathrm{E}((1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))^{N_{p}})\\
=\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{1-(1-p)(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\
=\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{1-(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\
 =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{-i\sqrt{p}t\mathrm{E}X+\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\
 =\frac{p(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}{p(-i\frac{\sqrt{p}}{p}t\mathrm{E}X+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})))}\\
=\frac{1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})}{-i\frac{\sqrt{p}}{p}t\mathrm{E}X+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t\mathrm{E}X-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\\
\end{split}
  \end{equation*}
Since
\begin{equation*}
\begin{split}
Z\in L(0,\sigma)\\
U=Z+\mu\\
\psi_{U}(t)=\psi_{Z+\mu}(t)=\mathrm{E}(e^{t(Z+\mu)})=\mathrm{E}(e^{\mu t} \cdot e^{tZ})=e^{\mu t}\mathrm{E}(e^{tZ})=e^{\mu t}\cdot \frac{1}{1-\frac{\sigma^{2}}{2}t^{2}}\\
U \in L(\mu,\sigma)
 \end{split}
  \end{equation*}
\begin{equation*}
\begin{split}
\mathrm{E}(X)=0
 \end{split}
  \end{equation*}
and
\begin{equation*}
\begin{split}
  \lim_{p\rightarrow 0} \varphi_{\sqrt{p}Y_{p}}(t)= \lim_{p\rightarrow 0}g_{N_{p}}( \varphi_{X}(\sqrt{p} t))\\
=\lim_{p\rightarrow 0}e^{\mu t}\cdot \big(\frac{1+i\sqrt{p}t(0)-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})}{-i\frac{\sqrt{p}}{p}t(0)+\frac{t^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2})+(1+i\sqrt{p}t(0)-\frac{pt^{2}}{2}\sigma^{2}+\mathscr{o}(pt^{2}))}\big)\\
=e^{\mu t}\cdot \frac{1}{1-\frac{\sigma^{2}}{2}t^{2}}\\
\sqrt{p}Y_{p}\rightarrow L(\sigma)\quad \text{as} \quad p\rightarrow 0.
\end{split}
  \end{equation*}
