Question: > 
> Let $\{X\_n\}\_{n\ge1}$ be a sequence of i.i.d. random variables with common mean $\mu$ and variance $\sigma^2 \in (0,\infty)$. Use Slutsky's theorem to show that:
> \begin{align}
> \sqrt{n}(e^{\frac{S\_n}{n}}-e^{\mu}) \xrightarrow{d} \sigma e^{\mu}Z
> \end{align}
> where $S\_n=\sum\limits\_{k=1}^{n}X\_k$ and $Z\in N(0,1)$.
> 
> 
> 

I have used Slutsky's theorem in plenty of problems before but I cannot make any progress on this one, any help here would be greatly appreciated.

Answer: 
Define
$$
G(t)=\begin{cases}\dfrac{e^t-\mu}{t-\mu}, & t\neq \mu,\cr e^\mu, & t=\mu.\end{cases}
$$
This function is continuous everywhere since $\lim\limits\_{t\to\mu}\dfrac{e^t-\mu}{t-\mu}=(e^t)'\_\mu=e^\mu$. By LLN, $\frac{S\_n}{n}\xrightarrow{p}\mu$, and by continuous mapping theorem,
$$
G\left(\frac{S\_n}{n}\right)\xrightarrow{p}G(\mu)=e^\mu.
$$
Then
$$
\sqrt{n}\left(e^{\frac{S\_n}{n}}-e^{\mu}\right)=\sqrt{n}\left(\frac{S\_n}n-\mu\right)\cdot G\left(\frac{S\_n}{n}\right).
$$
Here the first term $\sqrt{n}\left(\frac{S\_n}n-\mu\right)\xrightarrow{d}\sigma Z$ by CLT, where $Z\sim N(0,1)$. And the second term converges in probability to $e^\mu$. Slutsky's theorem implies that the product converges in distribution to $\sigma e^\mu Z$.

Note that this is almost the same reasoning as in the previous answer, and indeed the same as the so-called delta method.
this is not a direct solution to your problem, but this situation seems a candidate for an application of the delta method. Here is the result. I think filling the details for your situation is straight forward:

**Theorem:** Let $X\_n$, $n\in\mathbb{N}$, and $Y$ be random vectors in
$\mathbb{R}^m$. Suppose $a\_n\xrightarrow{n\rightarrow\infty}\infty$ and
$a\_n(X\_n-c)\stackrel{n\rightarrow\infty}{\Longrightarrow} Y$. If $g$ is a differentiable function, then
\begin{aligned}
a\_n(g(X\_n)-g(c))\stackrel{n\rightarrow\infty}{\Longrightarrow} g'(c)Y
\end{aligned}

You can find this result in many statistics books (Shao's Mathematical Statistics for example) The proof combines a first order Taylor approximation of $g(X\_n)$ and makes use of Slutky's lemma.

Here is a **sketch of the proof:**
Using a first order Taylor expansion we obtain
\begin{equation}
g(x)-g(c)-g'(c)(x-c)=o(x-c)
\end{equation}
where $|o(x-c)|/|x-c|\rightarrow0$ as $x\rightarrow c$. By Slutsky's theorem and convergence in measure, it is enough to show that $a\_no(X\_n-c)$ converges in law, an hence in measure, to $0$.
Since $a\_n(X\_n-c)$ converges in law and $a\_n\rightarrow\infty$. Slutsky's theorem and convergence in measure arguments show that
$X\_n-c=\frac{1}{a\_n}a\_n(X\_n-c)$ converges in law, and hence in measure, to $0$. For any $\varepsilon>0$, there is $\delta>0$ such that $|x-c|<\delta$ implies that
$|o(x-c)|<\varepsilon|x-c|$. Hence,
\begin{aligned}
\Pr[|o(X\_n-c)|/|X\_n-c|\geq\varepsilon]\leq\Pr[|X\_n-c|\geq\delta]\rightarrow0
\end{aligned}
as $n\rightarrow\infty$, and so $o(X\_n-c)/|X\_n-c|$ converges in measure to $0$.
As $a\_n(X\_n-c)$ converges in law, by Slutsky's theorem
$|a\_no(X\_n-c)|=|a\_n(X\_n-c)|\frac{|o(X\_n-c)|}{|X\_n-c|}$ converges in law, and hence in measure, to $0$.