The question states: Let $X_1,....X_n$ be a random sample from $f(x$,$\theta$) with $E(X$) = $\mu$ and $V(X) = \sigma^2$. Find the limiting distribution of $\bar{X}^3$, and the approximate distribution of $\bar{X}^3$. So I use the delta method to show that $\sqrt{n}(\bar{X} - \mu)$ $\overset{D}{\rightarrow}$ $N(0,\sigma^2)$ then $\sqrt{n}(\bar{X}^3 - \mu^3)$ $\overset{D}{\rightarrow}$ $N(0,9\sigma^2\mu^4)$ for large n. Through taking the expected value and variance of the left side I get $\bar{X}^3$ is approximately $N(\mu^3,\frac{9\sigma^2\mu^4}{n})$. In this type of situation is the limiting distribution and approximate distribution of $\bar{X}^3$ the same thing?
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"Limiting" and "approximate" are not necessarily the same thing. Limiting is specifically a distribution coming from some asymptotic result, so if you take $Y=t(\underline{X})$ and let's say that $\mu_Y= \theta(1-4/n)$ and $\sigma_Y=(2+2/n)/\sqrt{n}$ and you can show that $\lim_{n\to\infty}\frac{Y-\theta}{2/\sqrt{n}}\sim N(0,1)$, you can say "the limiting distribution of $\sqrt{n}(Y-\theta)$ is $N(0,2^2)$", and you might from that obtain an approximation for the distribution of $Y$ at $n=8$, say but for $n=8$ the distribution of $Y$ may be better approximated by a normal with $\mu=\frac{\theta}{2}$ and $\sigma^2=81/128$ (if I did that right) than what you'd get from substituting $n$ into the limit calculation.
So "are they the same" in general depends on what approximation you're using.
If you used a limit result for a standardized $Y$ to obtain your approximation for for the distribution of $Y$ at some $n$ (which it looks like you did) then you might loosely(!) say that you're using the limiting distribution, though really it's an approximation obtained from the limiting distribution.
