Given an IID sample $X_1, \dots, X_n \sim N(\mu, \sigma^2)$ with $\mu \ne 0$ let
$$(S_n)^2 = \frac{1}{n} \sum_{i = 1}^n ( X_i - \overline{X_n})^2$$
I want to find the distribution of the coefficient of variation $\frac{S_n}{\bar{X}_n}$ where $\bar{X}_n$ is the sample mean. I have come up with an entire solution but there appears to be an error I can't find.
MY SOLUTION:
$S_n$ is asymptotically consistent for $\sigma \implies \frac{S_n}{\sigma} \xrightarrow{p}1$
Since $ \sqrt{n}(\bar{X}_n - \mu)$ converges to a normal distribution (by the central limit theorem) by the delta method we have that $\sqrt{n} \frac{\sigma}{\bar{X}_n}$ converges in distribution to $$ \frac{\sigma^2}{\mu^2}Z $$
where $Z \sim N\left(\frac{\sigma}{\mu}, 1\right) $ this is because we have chosen $\phi(x) = \frac{\sigma}{x} \implies \phi'(x) = -\frac{\sigma}{x^2} $. ($\phi$ is the function we are applying to the sequence of random variables).
Now from Slutsky's theorem we have that $$\sqrt{n} \frac{S_n}{\bar{X}_n } = \sqrt{n}\frac{S_n/ \sigma}{\bar{X}_n / \sigma} \xrightarrow{d} N\left( \frac{\sigma}{\mu},\frac{\sigma^4}{\mu^4}\right)$$
Except the solution I am given has $N\left(\frac{\sigma}{\mu},\frac{\sigma^2}{2\mu^2}\right)$
Where is my mistake?
First of all, $\sqrt{n} \frac{S_n}{\bar{X}_n }$ diverges in distribution. Since the fraction tends to $\frac{\sigma}{\mu}$ in probability, and $\sqrt{n}$ tends to infinity, so the product tends to $\pm\infty$ as well.
I suppose that you want to find the limit in distribution of $$\sqrt{n}\left(\frac{S_n}{\bar{X}_n } - \frac{\sigma}{\mu}\right).$$
So, $$\tag{1}\label{1} \sqrt{n}\left(\frac{S_n}{\bar{X}_n } - \frac{\sigma}{\mu}\right)=\frac{1}{\bar{X}_n}\sqrt{n}\left(S_n - \frac{\sigma}{\mu}\bar{X}_n\right)=\frac{1}{\bar{X}_n}\left(\sqrt{n}\left(S_n - \sigma\right) - \sqrt{n}\left(\frac{\sigma}{\mu}\bar{X}_n-\sigma\right)\right).$$
Note that $$ \sqrt{n}\left(\frac{\sigma}{\mu}\bar{X}_n-\sigma\right)\xrightarrow{d} N\left(0,\frac{\sigma^4}{\mu^2}\right). $$ In fact, this value is normally distributed for any $n$, but it does not matter now.
Next, $$ \sqrt{n}\left(S_n - \sigma\right)=\frac{\sqrt{n}\left(S^2_n - \sigma^2\right)}{S_n+\sigma} $$ For the denominator we have $S_n+\sigma \xrightarrow{p} 2\sigma$, and for the numerator the following holds: $\sqrt{n}\left(S^2_n - \sigma^2\right) \xrightarrow{d} N\left(0,2\sigma^4\right)$. Then by Slutsky's theorem we get $$ \sqrt{n}\left(S_n - \sigma\right)=\frac{\sqrt{n}\left(S^2_n - \sigma^2\right)}{S_n+\sigma} \xrightarrow{d} N\left(0,\frac{2\sigma^4}{(2\sigma)^2}\right) = N\left(0,\frac{\sigma^2}{2}\right). $$ Note that $S_n$ and $\bar X_n$ are independent. Therefore the difference in (\ref{1}) converges in distribution to a difference of two independent normally distributed random variables: $$ \sqrt{n}\left(S_n - \sigma\right) - \sqrt{n}\left(\frac{\sigma}{\mu}\bar{X}_n-\sigma\right) \xrightarrow{d} N\left(0,\frac{\sigma^2}{2}+\frac{\sigma^4}{\mu^2}\right) $$ Finally, by Slutsky's theorem with $\bar X_n\xrightarrow{p}\mu$, we get: $$ \sqrt{n}\left(\frac{S_n}{\bar{X}_n } - \frac{\sigma}{\mu}\right)=\frac{1}{\bar{X}_n}\left(\sqrt{n}\left(S_n - \sigma\right) - \sqrt{n}\left(\frac{\sigma}{\mu}\bar{X}_n-\sigma\right)\right)\xrightarrow{d} N\left(0,\frac{\sigma^2}{2\mu^2}+\frac{\sigma^4}{\mu^4}\right).$$