Hint:

let we know by central limit theorem $$\sqrt{n}(\bar{X}-\mu_{x})\rightarrow N(0,\sigma^2)$$

By using Taylor series we get $$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)=\sqrt{n} (\bar{X}-\mu_{x})h^{\prime}(\mu_{x}) +\sqrt{n} (\bar{X}-\mu_{x})^2\frac{h^{\prime \prime}(\mu_{x})}{2!}+\cdots$$

lets we use an approximation like (if $h^{\prime}(\mu_{x})\neq 0$)

$$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)\cong \color{red}{\sqrt{n} (\bar{X}-\mu_{x})}h^{\prime}(\mu_{x})\rightarrow N(0,?) $$

Hint:

let we know by central limit theorem $$\sqrt{n}(\bar{X}-\mu_{x})\rightarrow N(0,\sigma^2)$$

By using Taylor series we get $$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)=\sqrt{n} (\bar{X}-\mu_{x})h^{\prime}(\mu_{x}) +\sqrt{n} (\bar{X}-\mu_{x})^2\frac{h^{\prime \prime}(\mu_{x})}{2!}+\cdots$$

lets we use an approximation like (if $h^{\prime}(\mu_{x})\neq 0$)

$$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)\cong \color{red}{\sqrt{n} (\bar{X}-\mu_{x})}h^{\prime}(\mu_{x})\rightarrow N(0,?) $$

Hint:

let we know by central limit theorem $$\sqrt{n}(\bar{X}-\mu_{x})\rightarrow N(0,\sigma^2)$$

By using Taylor series we get $$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)=\sqrt{n} (\bar{X}-\mu_{x})h^{\prime}(\mu_{x}) +\sqrt{n} (\bar{X}-\mu_{x})^2\frac{h^{\prime \prime}(\mu_{x})}{2!}+\cdots$$

Hint:

let we know by central limit theorem $$\sqrt{n}(\bar{X}-\mu_{x})\rightarrow N(0,\sigma^2)$$

By using Taylor series we get $$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)=\sqrt{n} (\bar{X}-\mu_{x})h^{\prime}(\mu_{x}) +\sqrt{n} (\bar{X}-\mu_{x})^2\frac{h^{\prime \prime}(\mu_{x})}{2!}+\cdots$$

lets we use an approximation like (if $h^{\prime}(\mu_{x})\neq 0$)

$$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)\cong \color{red}{\sqrt{n} (\bar{X}-\mu_{x})}h^{\prime}(\mu_{x})\rightarrow N(0,?) $$