The asymptotic distributions of $X^2$, $\frac{1}{X}$ and $\exp(X)$ are obtained as follows:X²The asymptotic distribution of $X^2$ is obtained by the delta method. If $g(x) = x^2$ and $\mu = E(X)$ and $\sigma^2 = Var(X)$, then$$ \sqrt{n}(g(\bar X_n) - g(\mu)) \to_d N(0,\sigma^2(g'(\mu))^2) $$Hence, $$ \sqrt{n}(X^2 - \mu^2) \to_d N(0,4\mu^2 \sigma^2) $$Note that we have used the delta method twice: once for the square and once for the mean. 1/XSuppose $g(x) = \frac{1}{x}$. Then $g'(x) = -\frac{1}{x^2}$. By the delta method,$$ \sqrt{n}(g(\bar X_n) - g(\mu)) \to_d N(0,\sigma^2(g'(\mu))^2) $$Hence,$$ \sqrt{n}(\frac{1}{\bar X_n} - \frac{1}{\mu}) \to_d N(0,\frac{\sigma^2}{\mu^4}) $$exp(X)If $g(x) = \exp(x)$, then $g'(x) = \exp(x)$. By the delta method,$$ \sqrt{n}(g(\bar X_n) - g(\mu)) \to_d N(0,\sigma^2(g'(\mu))^2) $$Hence, $$ \sqrt{n}(\exp(\bar X_n) - \exp(\mu)) \to_d N(0,\exp(2\mu)\sigma^2) $$Note that we have used the delta method twice: once for the exponential and once for the mean. The assumptions made are that the random sample is drawn from a population with finite mean and variance.