Find the limiting distribution of the sequence The following is a qualifying exam problem. 

Let $\{X_n\}$ be an i.i.d. sequence of Binomial($n,1/2$) random variables. Define the sequence, 
$$ Y_n = \left(1+\frac{1}{\sqrt{n}}\right)^{X_n}\left(1-\frac{1}{\sqrt{n}}\right)^{n-X_n} $$
Find the limiting distribution of $Z_n = \ln(Y_n)$. 

I tried using a characteristic function approach. The chf of $Z_n$ is given by, 
$$ \begin{eqnarray*}\phi_{Z_n}(t) &=& E[e^{itZ_n}] \\ &=& E[e^{it\ln(Y_n)}] \\ 
&=& E[e^{itX_n\ln\left(1+\frac{1}{\sqrt{n}}\right)}e^{it(n-X_n)\ln\left(1-\frac{1}{\sqrt{n}}\right)}]\\
&=& \sum\limits_{k=0}^ne^{it\ln\left(1+\frac{1}{\sqrt{n}}\right)k}e^{it\ln\left(1-\frac{1}{\sqrt{n}}\right)(n-k)}\binom{n}{k}\frac{1}{2^n}\\
&=& \left(\frac{(1+\frac{1}{\sqrt{n}})^{it}+(1-\frac{1}{\sqrt{n}})^{it}}{2}\right)^n \end{eqnarray*} $$
A computation in Mathematica reveals that, 
$$ \lim\limits_{n\rightarrow\infty}\phi_{Z_n}(t) = e^{-\frac{1}{2}it - \frac{1}{2}t^2} $$
which we recognize as the chf of a Gaussian distribution with mean $-\frac{1}{2}$ and variance $1$. 
A couple questions: 
1) Is there a straightforward way of computing the limit above (without the use of CAS software)?
2) Is there a more time efficient approach here? Keep in mind that this was a qualifying exam problem. 
 A: You can use the central limit theorem:
\begin{align}
&\ln Y_n \\
& = X_n \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}} + n \ln \left(1-\frac{1}{\sqrt{n}}\right) \\
% 
% & = \underbrace{\sqrt{n} \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}}_{\xrightarrow[n\to\infty]{} \, 2} \times \sqrt{n}\left( \frac{X_n}{n} - \frac{1}{2} \right)
% + \frac{1}{2}n \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}
% + n \ln \left(1-\frac{1}{\sqrt{n}}\right) \\
% 
% 
% 
& = \sqrt{\mbox{Var}(X_1)} \underbrace{\sqrt{n} \ln \frac{1+\frac{1}{\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}}_{\xrightarrow[n\to\infty]{} \, 2} \times \frac{\sqrt{n}}{\sqrt{\mbox{Var}(X_1)}}\left( \frac{X_n}{n} - \frac{1}{2} \right) +
\frac{n}{2} \ln \left( 1+\frac{1}{\sqrt{n}}\right) + \frac{n}{2} \ln \left( 1-\frac{1}{\sqrt{n}}\right) \\
% 
% 
% 
&\xrightarrow[n\to\infty]{D} Z-\frac{1}{2}
\end{align}
where Var$(X_1)=1/4$, $Z\sim N(0,1)$, and in the last I used that
\begin{align}
\lim_{n\to\infty} \left( \left( 1+\frac{1}{\sqrt{n}}\right)\left( 1-\frac{1}{\sqrt{n}}\right)\right)^n
= \lim_{n\to\infty} \left( 1-\frac{1}{n}\right)^n = \frac{1}{e}
\end{align}
