Find the limiting distribution of $\sqrt{n} \left(\sqrt{\bar{X}} -1 \right) $ if $\sqrt{n} \left( \bar{X}-1 \right) \to N(0,1)$ Find the limiting distribution of $\sqrt{n} \left(\sqrt{\bar{X}} -1 \right) $ if $\sqrt{n} \left( \bar{X}-1 \right) \to N(0,1)$. Can you please check my work below?
In principle, the Delta method should be of use here. We know that according to that particular method if $$ \sqrt{n} \left( X_n -\theta \right) \to N \left( 0,\sigma^2 \right)$$ then $$\sqrt{n} \left( g\left(X_n \right) -g \left(\theta \right) \right) \to N\left( 0,\sigma^2 g \prime \left(\theta \right)^2 \right)$$
We have $g \left(t \right) =\sqrt{t}$ and $g(1)=1$ which implies that $g \prime \left( 1 \right)= \frac{1}{2}$ and then the limiting distribution is $N \left(0, 1/4 \right)$
Is everything alright here? Thanks.
 A: The result is correct (up to a factor of $\sigma$, which is an unimportant typographical omission).  This answer provides two separate ways to double-check it.
We can in fact obtain the PDF of the transformed variables directly: when the $X_n$ are exactly Normal (and not just asymptotically so), the PDF of $\sqrt{n}\left(g(X_n)-1\right)$ can be found via integration as
$$\frac{1}{(\sigma/2)\sqrt{2\pi n}}  \left(\sqrt{n}+x\right) \exp\left({-\frac{x^2 \left(2 \sqrt{n}+x\right)^2}{2 n \sigma ^2}}\right)$$
for $x\gt -\sqrt{n}$ (and equal to $0$ otherwise).  For fixed $x$, the limiting value as $n\to\infty$ is 
$$\frac{1}{(\sigma/2)\sqrt{2\pi}}  \exp\left({-\frac {8x^2}{2 \sigma ^2}}\right) = \frac{1}{(\sigma/2)\sqrt{2\pi}}  \exp\left({-\frac {x^2}{2 (\sigma/2) ^2}}\right),$$
the PDF of a Normal$(0, \sigma^2/4)$ distribution.

One can check also check the result with a simulation, such as carried out by this R code:
set.seed(17)
n <- 10^6
x <- sqrt(n)*(sqrt(rnorm(n, 1, 1/sqrt(n)))-1)
m <- mean(x); v <- var(x); k <- mean((x-m)^4)
se.v <- sqrt(((n-1)^2 * k - (n-1)*(n-3)*v^2) / n^3)
print(v)              # Variance
print((v - 1/4)/se.v) # Its standardized standard error

This reports a simulated variance (when $n=10^6$) of $0.25015$ times the (unit) standard deviation, which is just $0.44$ standard errors away from $1/4$: evidence that the computed variance is not incorrect for such a large $n$.
