Exchanging limits when functions may not converge uniformly 
Let $X$ be a compact metric space and for each $i \geq 1$, let $f_i \colon X \to \mathbb{N}$ be continuous functions satisfying:
  
  
*
  
*$f_{i+1}(x) \geq f_i(x)$ and;
  
*for each $n \in \mathbb{N}$ there exists an $x \in X$ and  $j\geq 1$ such that $f_j(x) \geq n$.
  
  
  Does there then exist $x \in X$ such that ${\displaystyle \lim_{i \to \infty}(f_i(x)) = \infty}$?

The above situation appeared in my research on metric dynamical systems (in a more specific setting, but I think the above criteria is the relevant information). My attempt was to take the sequence of elements $x_n$ and $j_n$ such that $f_{j_n}(x_n) \geq n$ and then form a convergent subsequence $x_{n_k}$. The limit ${\displaystyle x := \lim_{k \to \infty}x_{j_k}}$ is then a candidate for the point we need.
However, we end up getting that
$$\begin{array}{rcl}
\displaystyle \lim_{i \to \infty}(f_i(x)) & = & \displaystyle  \lim_{i \to \infty}(f_i(\lim_{k \to \infty}x_{j_k})) \\
& = & \displaystyle  \lim_{i \to \infty}\lim_{k \to \infty}f_i(x_{j_k})
\end{array}
$$
and it is not clear at all to me that we are able to exchange the two limits. If we can, then we end up with $\displaystyle  \lim_{i \to \infty}\lim_{k \to \infty}f_i(x_{j_k}) = \lim_{k \to \infty}\lim_{i \to \infty}f_i(x_{j_k})$ and as the sequence $(f_i(x_{j_k}))_{i \in \mathbb{N}}$ is eventually bounded below by $k$, it follows that $\displaystyle \lim_{i \to \infty}(f_i(x)) \geq \lim_{k \to \infty} k = \infty$. The Moore-Osgood Theorem would allow this exchange of limits if the $f_i$ converged uniformly, but I don't know if we can assume uniform convergence here.
So my problem is in being able to justify the exchange of these limits. Is this always possible in the above situation? Is the statement above true in general or do I require stronger conditions on the space/functions (for instance, it's automatic if $X$ is connected)?

I don't think it is true this general:
Consider $X=\{0\}\cup\{\frac {1}{n}:n\in\mathbb {N\}}$ and define $f_n(x)=\frac {1}{x}$ for $x\geq \frac {1}{n}$ and zero otherwise.
