Assumption to apply the delta method When proving the delta method of distributions in my textbook we make the following assumption:
Let $X_{n}$ be a sequence of random variables.
and:
${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,1)}$
Bearing this in mind it can be shown that:
${\sqrt{n}[g(X_n)-g(c)]\,\xrightarrow{D}\,\mathcal{N}(0,[g'(c)]^2)}$
What I'm struggling to understand why in this case the assumption to implement the delta method is:
${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,1)}$
and not
${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^{2})}$?
Usually $X_{n}$ is an estimator of a population parameter and $c$ the actual parameter being estimated. If the estimator is a consistent asymptotically normal estimator we can conclude that:
${\frac{\sqrt{n}[X_n- c]}{\sigma}\,\xrightarrow{D}\,\mathcal{N}(0,1)}$
Would it be safe to assume that here $\sigma^{2} = 1$ is assumed here?

The two statements :
$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, 1)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2)\right) $$
and
$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2\sigma^2)\right) $$
are equivalent.
The second one clearly implies the first one (take $\sigma = 1$).
Now, assume the first one is true, and suppose that
$$\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2). $$
Define $Y_n = \frac{X_n}{\sigma}$, $c' = \frac{c}{\sigma}$ and $g_1 : t \mapsto g(\sigma t)$.
Then you have
$$\sqrt{n}(Y_n - c') \overset{D}{\to}\mathcal{N}(0, 1),$$
and thus
$$\sqrt{n}\left(g_1(Y_n) - g_1(c')\right) \overset{D}{\to}\mathcal{N}(0, g_1'(c')^2)$$
which gives
$$\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2 \sigma^2).$$
