On the existence of a non-negative function on a Banach space whose limit at every point is infinity. Does there exist a Banach space $ X $ (possibly non-separable) and a mapping $ F: X \to X $ such that
$$
\forall a \in X: \quad
\lim_{\substack{x \in X \setminus \{ a \} \\ x \to a}} \| F(x) \|_{X} = \infty?
$$
Note: If the Banach space $ X $ is trivial, then the answer is a vacuous affirmative as the zero element, $ 0_{X} $, is the sole element of $ X $ and hence not a limit point. We may thus restrict our attention to only non-trivial Banach spaces.
 A: The answer is ‘no’. In fact, we can derive the statement from the following theorem:

Theorem. Let $ X $ be a non-empty Baire space where every point is a limit point. Then there is no function $ f: X \to \mathbb{R}_{\geq 0} $ such that
  $$
\lim_{\substack{x \in X \setminus \{ a \} \\ x \to a}} f(x) = \infty
$$
  for every $ a \in X $.

Proof
Assume that the theorem is false, i.e., assume that there exists a function $ f: X \to \mathbb{R}_{\geq 0} $ such that
$$
\lim_{\substack{x \in X \setminus \{ a \} \\ x \to a}} f(x) = \infty
$$
for every $ a \in X $.
Define a sequence $ (C_{n})_{n \in \mathbb{N}} $ of subsets of $ X $ by
$$
\forall n \in \mathbb{N}: \quad
C_{n} \stackrel{\text{df}}{=} \{ x \in X \mid f(x) \in [0,n] \}.
$$
We claim that $ C_{n} $ is a closed subset of $ X $ for each $ n \in \mathbb{N} $. Let $ n \in \mathbb{N} $, and suppose that $ a \notin C_{n} $. As
$$
\lim_{\substack{x \in X \setminus \{ a \} \\ x \to a}} f(x) = \infty,
$$
we can find an open neighborhood $ U $ of $ a $ such that $ f(x) > n $ for all $ x \in U \setminus \{ a \} $. It follows that $ U \cap C_{n} = \varnothing $. Therefore, $ X \setminus C_{n} $ is a union of open subsets of $ X $, which implies that $ C_{n} $ is indeed a closed subset of $ X $.
Now, $ \displaystyle X = \bigcup_{n = 1}^{\infty} C_{n} $, and as it is a non-empty Baire space by assumption, there exists an $ N \in \mathbb{N} $ such that $ C_{N} $ has a non-empty interior $ E $. Choose a $ b \in E $, and let $ V $ be an open neighborhood of $ b $ such that $ f(x) > N $ for all $ x \in V \setminus \{ b \} $. However, $ E \cap V $ is an open neighborhood of $ b $, and we already know that $ b $ is a limit point of $ X $. Hence, $ (E \cap V) \setminus \{ b \} \neq \varnothing $, which is a clear contradiction as there now exists an $ x \in (E \cap V) \setminus \{ b \} $ such that $ f(x) \leq N $ on one hand and $ f(x) > N $ on the other.
Our initial assumption that $ f $ exists is thus false, and so the theorem is established. $ \quad \blacksquare $

To prove the original statement, simply observe that a non-trivial Banach space $ X $ is a Baire space (by the Baire Category Theorem) where every point is a limit point. Then define $ f: X \to \mathbb{R}_{\geq 0} $ by
$$
\forall x \in X: \quad
f(x) \stackrel{\text{df}}{=} \| F(x) \|_{X}.
$$
We ignore the trivial Banach space because the zero element, the sole element of the space, is not a limit point.
A: No. Since the constant sequence $x_i = a$ converges to $a$ this would mean $\|F(a)\| = +\infty$.
However, as Berci suggested, you can ask for $f : X \to \mathbb R$, such that
$$\limsup_{x \to a} f(x) = +\infty.$$
This is easy to accomplish, even with $X = \mathbb R$. Enumerate the rationals $\mathbb Q = \{q_i\}$ and define
$$f(x) = \begin{cases} i & \text{if }x = q_i \text{ for some } i,\\0 & \text{if } x \not\in \mathbb Q.\end{cases}$$
This should do the job.
