= Solution
Fix an interior point $o\in\mathbb H^3$. The <limit set of a Kleinian group> is the set of boundary accumulation points of its orbit, or
$$
\boxed{\Lambda(\Gamma)=\overline{\Gamma o}^{\,\mathbb H^3\cup S^2_\infty}\cap S^2_\infty.}
$$
The boundary compactification may be realized as the closed unit ball. A discrete isometry group has only finitely many orbit points in each compact interior region: isometries moving $o$ into a compact set form a compact family after accounting for the compact rotation stabilizer, and infinitely many distinct group elements there would contradict discreteness. The stabilizer of $o$ is finite for the same reason. Thus an infinite group has an infinite orbit escaping every compact interior set and a nonempty boundary accumulation set. A finite group has empty limit set. This also shows the displayed intersection does not introduce ordinary interior accumulation points.
For another point $o'$, the corresponding points satisfy $d(\gamma o,\gamma o')=d(o,o')=:C$. In the Poincaré ball model the distance identity is
$$
\sinh\frac{d(u,v)}2=\frac{|u-v|}{\sqrt{(1-|u|^2)(1-|v|^2)}}.
$$
If $\gamma_j o\to\xi\in S^2_\infty$, bounded distance implies
$$
|\gamma_j o-\gamma_j o'|
\le\sinh(C/2)\sqrt{(1-|\gamma_j o|^2)(1-|\gamma_j o'|^2)}\longrightarrow0.
$$
So $\gamma_j o'\to\xi$ as well. Exchanging $o,o'$ proves equality of the two accumulation sets. Hence the definition is \b[independent of the interior base point].
It is closed as an intersection of the compact orbit closure with the closed boundary, and it is $\Gamma$-invariant because multiplying an orbit by an element of $\Gamma$ permutes that orbit. Changing ball coordinates or conjugating the geometric realization transports the set by the induced boundary <homeomorphism>. Different nonconjugate representations of an abstract group are not asserted to have the same subset of one fixed sphere. None of these arguments requires finite generation.
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