Answer
$$f^{-1}(x)=\sqrt[3]{x+4}$$
$$\lim_{x \to 4 }f^{-1}(x)=2$$
Work Step by Step
$$y=x^3-4 \quad \to \quad x=\sqrt[3]{y+4} \\ \Rightarrow \quad f^{-1}(x)=\sqrt[3]{x+4}$$
To find $\lim_{x\to 4 }f^{-1}(x)$, examine the graph of $f^{-1}$ near $x=4$. As $x$ gets closer to $4$, the values of $f^{-1}(x)$ get closer to $2$. We conclude from the graph that$$\lim_{x \to 4 }f^{-1}(x)=2.$$
