Asymptotic distribution of the $t$-statistic Let $X_i$ be i.i.d. variables and $\hat{\mu}$, $\hat{\sigma}$ their sample mean and standard deviation.
Is the following true:
$$ \sqrt{n} \left( \frac{\hat{\mu}}{\hat{\sigma}} - \frac{\mu}{\sigma} \right) \to N(0, 1)?$$
I have tried to prove it using continuity/Slutsky/delta style arguments, but I can't seem to figure it out.
Here are two similar results I know how to prove:
$$ \sqrt{n} \left( \frac{\hat{\mu}- \mu}{\sigma} \right) \to N(0, 1)$$
$$ \sqrt{n} \left( \frac{\hat{\mu}- \mu}{\hat{\sigma}} \right) \to N(0, 1)$$
The first is just the CLT. I found a proof of the second one online: http://www.datascienceassn.org/sites/default/files/Asymptotic%20Statistics%20-%20Lecture%20Notes.pdf (see example 1.2.8)
 A: $\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}$
$\DeclareMathOperator*{\Var}{Var}
\DeclareMathOperator*{\Cov}{Cov}
\newcommand{\gaussian}{\mathcal{N}}$
To apply multivariate CLT and Delta method to derive the asymptotic distribution, you need to impose additional conditions that quantifies the variation of $X_1^2$ and the correlation between $X_1$ and $X_1^2$.  In general, the asymptotic normality still holds, but the asymptotic variance is not $1$.  Below is a derivation of getting the asymptotic distribution of $\hat{\mu}/\hat{\sigma}$ when $X_1, \ldots, X_n \text{ i.i.d.} \sim \gaussian(\mu, \sigma^2)$, which disproves your conjecture as long as $\mu \neq 0$.  The proof can be easily generalized to other underlying distributions (e.g., uniform, Poisson, etc.), or when $\Var(X_1^2)$ and $\Cov(X_1, X_1^2)$ are given as known values (in terms of distributional parameters).
For notational convenience, denote $n^{-1}\sum\limits_{i = 1}^n X_i^k$ by $\overbar{X_n^k}, k = 1, 2$, $n^{-1}\sum\limits_{i = 1}^n(X_i - \overbar{X_n})^2$ by $S_n^2$.  Direct evaluation yields:
\begin{align}
& E[X_1] = \mu, E[X_1^2] = \mu^2 + \sigma^2, \\
& \Var(X_1) = \sigma^2, \Var(X_1^2) = 2\sigma^4 + 4\mu^2\sigma^2, \\
& \Cov(X_1, X_1^2) = 2\mu\sigma^2.
\end{align}
It then follows by the multivariate CLT that
\begin{equation*}
\sqrt{n}\left[\begin{pmatrix} \overbar{X_n} \\ \overbar{X_n^2} \end{pmatrix} - \begin{pmatrix} \mu \\ \mu^2 + \sigma^2 \end{pmatrix}\right] \Rightarrow 
\gaussian\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} \sigma^2 & 2\mu\sigma^2 \\
2\mu\sigma^2 & 2\sigma^4 + 4\mu^2\sigma^2 \end{pmatrix}\right).
\end{equation*}
To facilitate the Delta method, define
$$g(x, y) = \frac{x}{\sqrt{y - x^2}}, \; (x, y) \in D = \{(x, y) \in \mathbb{R}^2: y \neq x^2\}.$$
It is easy to verify that $g$ is differentiable everywhere on $D$, and
\begin{equation*}
\nabla g(x, y) = 
\begin{pmatrix} \dfrac{\partial g(x, y)}{\partial x} \\ 
\dfrac{\partial g(x, y)}{\partial y}  \end{pmatrix}
= \begin{pmatrix} \dfrac{y}{(y - x^2)^{3/2}} \\ 
-\dfrac{x}{2(y - x^2)^{3/2}} \end{pmatrix}.
\end{equation*}
Under the condition $\sigma^2 > 0$, $\nabla g(\mu, \mu^2 + \sigma^2)$ is well-defined. Furthermore,
$$g(\overbar{X_n}, \overbar{X_n^2}) = \dfrac{\overbar{X_n}}{S_n}, \;
g(\mu, \mu^2 + \sigma^2) = \dfrac{\mu}{\sigma},$$
and
\begin{align*}
& \nabla^T g(\mu, \mu^2 + \sigma^2)  \begin{pmatrix} \sigma^2 & 2\mu\sigma^2 \\
2\mu\sigma^2 & 2\sigma^4 + 4\mu^2\sigma^2 \end{pmatrix} \nabla g(\mu, \mu^2 + \sigma^2) \\
= & \frac{1}{\sigma^6} \begin{pmatrix} \mu^2 + \sigma^2 & -\dfrac{1}{2}\mu
\end{pmatrix} 
\begin{pmatrix} \sigma^2 & 2\mu\sigma^2 \\
2\mu\sigma^2 & 2\sigma^4 + 4\mu^2\sigma^2 \end{pmatrix} 
\begin{pmatrix} \mu^2 + \sigma^2 \\ -\dfrac{1}{2}\mu \end{pmatrix} \\
= & \frac{\mu^2}{2\sigma^2} + 1.
\end{align*}
Therefore,
\begin{equation*}
\sqrt{n}\left(\frac{\overbar{X_n}}{S_n} - \frac{\mu}{\sigma}\right) \Rightarrow \gaussian\left(0, \frac{\mu^2}{2\sigma^2} + 1\right).
\end{equation*}
Finally, note that $\hat{\mu} = \overbar{X_n}, \hat{\sigma}^2 = \frac{n}{n - 1}S_n^2$, whence by Slutsky's lemma:
\begin{align}
& \sqrt{n}\left(\frac{\hat{\mu}}{\hat{\sigma}} - \frac{\mu}{\sigma}\right) = \sqrt{n}\left(\sqrt{\frac{n - 1}{n}}\frac{\overbar{X_n}}{S_n} - \frac{\mu}{\sigma}\right) \\
=& \sqrt{\frac{n - 1}{n}}\times \sqrt{n}\left(\frac{\overbar{X_n}}{S_n} - \frac{\mu}{\sigma}\right) - \frac{\mu}{\sigma}\frac{1}{\sqrt{n} + \sqrt{n - 1}} \\
\Rightarrow & \gaussian\left(0, \frac{\mu^2}{2\sigma^2} + 1\right).
\end{align}
