Let's define $S_{n}=\sum_{k=1}^{n} X_{k}$. According to strong law of large numbers we have $\frac{S_{n}}{n} \rightarrow E(X_{1})=2$ a.s. It follows from that $\frac{S_{n}}{n} \rightarrow 2$ in distribution.
According to the continuity theorem $\lim_{n\to\infty} \varphi_{\frac{S_{n}}{n}}(t)=\varphi_{\delta(2)}(t), \forall t\in \mathbb{R}.$
Especially for $t=2$ we have:$$ \varphi_{\frac{S_{n}}{n}}(2)=\varphi_{S_{n}}(\frac{2}{n})=\varphi_{\sum_{k=1}^{n} X_{k}}(\frac{2}{n})=\prod_{k=1}^{n} \varphi_{X_{k}}(\frac{2}{n})=(\varphi_{X_{1}}(\frac{2}{n}))^{n}$$ and we can conclude $\lim_{n\to\infty} \varphi_{\frac{S_{n}}{n}}(2)=\lim_{n\to\infty}(\varphi_{X_{1}}(\frac{2}{n}))^{n}.$ Function $x\rightarrow \ln(x)$ is continuous so limit and function can commute. $$\ln(\lim_{n\to\infty}\varphi_{\frac{S_{n}}{n}}(2))=\ln(\lim_{n\to\infty}(\varphi_{X_{1}}(\frac{2}{n}))^{n})=\lim_{n\to\infty}\ln(\varphi_{X_{1}}(\frac{2}{n})^{n})=\lim_{n\to\infty}(n\ln(\varphi_{X_{1}}(\frac{2}{n})).$$ On the other hand, $$\ln(\lim_{n\to\infty}\varphi_{\frac{S_{n}}{n}}(2))=\ln(\varphi_{\delta(2)}(2))=\ln(e^{4\cdot i})=4\cdot i.$$ Final solution is $\lim_{n\to\infty}(n\ln(\varphi_{X_{1}}(\frac{2}{n})))=4\cdot i$. I hope that solution is correct. Every comment is welcome, thanks in advance.