Answer
$$f^{-1}(x)= \sqrt[3]{{x}+2}$$
$$\lim_{x \to 6 }f^{-1}(x)=\sqrt[3]{{6}+2}=2$$
Work Step by Step
$$y=x^3-2 \quad \to \quad x=\sqrt[3]{{y}+2} \\ \Rightarrow \quad f^{-1}(x)=\sqrt[3]{{x}+2}$$
To find $\lim_{x\to 6 }f^{-1}(x)$, examine the graph of $f^{-1}$ near $x=6$. As $x$ gets closer to $6$, the values of $f^{-1}(x)$ get closer to $\sqrt[3]{{6}+2}$. We conclude from the graph that$$\lim_{x \to 6 }f^{-1}(x)=\sqrt[3]{{6}+2}=2.$$
