Give a random variable $X$ if we have that \begin{align} \\\lim_{ k \to \infty}\left (E\left[\vert X \vert^k\right]\right)^{\frac{1}{k}} =\infty \end{align} Does this mean that $X$ has unbounded support?
Comment: The support is said to be unbounded if the distribution function $F$ of $|X|$ satisfies $F(x) <1 $ for all $x \in \mathbb{R}$.
Then I already answer your question in the comments. Note that $\mathbf{P}(|X| \in [0, c]) = 1$ is equivalent to $F(c) = 1$ (I am guessing $F$ is the distribution of $|X|$). Hence, $X$ is bounded iff $F$ has bounded support. But if $X$ is bounded by $c,$ $\mathbf{E}(|X|^k)^{\frac{1}{k}}$ is bounded by $c$ as well (in particular, the limit $\lim\limits_{k \to \infty} \left( \mathbf{E} \left( |X|^k \right) \right)^{\frac{1}{k}}$, wheter it exists or not, cannot be infinite). So, if the latter limit is infinite, $X$ is unbounded and $F$ is of unbounded support.