Are all bound states normalizeable? Following Griffiths eq. (2.91) on p. 52 one may define a bound state to be an energy eigenstate
$$H|E\rangle=E|E\rangle\tag{1}$$
with an energy being smaller than the potential far away from the origin in the sense
$$\lim_{x\rightarrow\pm\infty} V(x)>E.\tag{2}$$
Intuitively we want bound states $|E\rangle$ to be states with a finite spread in position space and to be normalizeable. This means the probability density $|\langle x|E\rangle|^2=|\psi_E(x)|^2$ should vanish at infinity. Is the requirement (2) enough to guarantee this? This Wikipedia article claims this.
Their reasoning seems flawed to me though: Rewriting eq. (1) in position basis
$$\frac{\hbar^2}{2m}\psi_E''(x)=\left(V(x)-E\right)\psi_E(x) \tag{3}$$
they directly deduce that $\psi_E(x)$ must decay exponentially for large $x$ as $V(x)-E\gt 0$ for all $x$ bigger than some $X$.
I fail so see why this is true.
So let me expand my question - What really are bound states by definition? Griffiths approach seems to be a too weak condition.
 A: TL;DR: OP has a point. Griffiths' bound state condition
$$E~<~ V(\pm\infty) \tag{2.91} $$
on p. 52 is incomplete. The solution $\psi$ to the 1D TISE will be a linear combination of exponentially growing and exponentially damped solutions in the asymptotic regions.
One needs to impose an extra condition to get rid of the non-normalizable exponentially growing solution. A heavy-handed (although standard) condition (which Griffiths presumably implicitly has in mind) would be to declare that $\psi$ should be normalizable from the onset. However, it turns out that it is enough to e.g. impose an asymptotic bound (C) on $\psi$, cf. the following Theorem.

Theorem. Let there be given the 1D TISE
$$\psi^{\prime\prime}(x)~=~\frac{2m}{\hbar^2}(V(x)-E)\psi(x). \tag{A}$$
If$^1$
$$ \exists k,K>0\forall |x|\geq K:~~ {\rm Re}\left[\frac{2m}{\hbar^2}(V(x)-E)\right]  ~\geq~ k^2,\tag{B}$$
and if $\psi$ is asymptotically bounded
$$ \exists c,K>0\forall |x|\geq K:~~ |\psi(x)|~\leq ~c,\tag{C} $$
then $\psi$ is exponentially damped:
$$\exists C>0 \forall |x|\geq K:~~ |\psi(x)| ~\leq~Ce^{-k|x|}, \tag{D} $$
and hence normalizable/square integrable.

Sketched proof of Theorem:

*

*From fairly mild assumptions about the potential $V$, we may assume that the solution $\psi$ is sufficiently many times differentiable, cf. e.g. my Phys.SE answer here.


*Let us use polar decomposition
$$ \psi(x)~=~R(x)e^{i\Theta(x)}. \tag{E} $$
Note that the function $R$ might not be differentiable at its zeros $R(x)=0$, although $R=|\psi|$ is still continuous everywhere.


*For $|x|\geq K$ away from the zeros $R(x)\neq 0$, we calculate:
$$\begin{align} \frac{R^{\prime\prime}(x)}{R(x)}~\geq~&\frac{R^{\prime\prime}(x)-R(x)\Theta^{\prime}(x)^2}{R(x)}\cr
~\stackrel{(E)}{=}~&{\rm Re}\frac{\psi^{\prime\prime}(x)}{\psi(x)}\cr
~\stackrel{(A)}{=}~&{\rm Re}\left[\frac{2m}{\hbar^2}(V(x)-E)\right]\cr
~\stackrel{(B)}{\geq}~& k^2. \end{align}
\tag{F} $$
So $R$ is concave upward, $R^{\prime\prime}>0$.


*It follows that $R$ can not have more than 1 zero in each of the asymptotic regions $]-\infty,-K]$ and $[K,\infty[$. By possibly choosing a bigger $K>0$, we may assume that there are no zeros in the asymptotic regions:
$$\forall |x|\geq K:~~ \psi(x)\neq 0. \tag{G}$$


*Together with the bounds (C), $R^{\prime\prime}>0$ means that
$$\begin{align} 
\forall x\geq K:~~ R^{\prime}(x)~\leq~& 0,\cr
\forall x\leq -K:~~ R^{\prime}(x)~\geq~& 0. \end{align}\tag{H}$$


*Let us define
$$S(x)~:=~\ln R(x). \tag{I} $$
Eq. (H) becomes
$$\begin{align} 
\forall x\geq K:~~ S^{\prime}(x)~\leq~& 0,\cr
\forall x\leq -K:~~ S^{\prime}(x)~\geq~& 0. \end{align}\tag{J}$$
Eq. (F) transforms into a Riccati inequality:
$$\forall |x|\geq K:~~ 
S^{\prime\prime}(x) +S^{\prime}(x)^2~\geq~ k^2.\tag{K} $$


*One may now show that
$$\begin{align} 
\forall x\geq K:~~ S^{\prime}(x)~\leq~& -k,\cr
\forall x\leq -K:~~ S^{\prime}(x)~\geq~& k. \end{align}\tag{L}$$
Indirect proof of eq. (L): Assume the opposite:
$$\begin{align} 
\exists x_1\geq K:~~ S^{\prime}(x_1)~>~& -k,\cr
\exists x_1\leq -K:~~ S^{\prime}(x_1)~<~& k. \end{align}\tag{M}$$
Then we may rewrite the Riccati ineq. (K) as
$$
\frac{1}{k}\frac{d}{dx}{\rm artanh}\frac{S^{\prime}(x)}{k} ~=~\frac{S^{\prime\prime}(x)}{k^2-S^{\prime}(x)^2} ~\stackrel{(K)}{\geq}~ 1\tag{N} $$
in a neighborhood of $x_1$. In particular we note that  $S^{\prime\prime}(x)>0$, i.e. $S^{\prime}$ is an increasing function. Therefore, in the 2 cases of eq. (M), we may let $x_1\to\pm\infty$, respectively, without violating eq. (M). Integration of ineq. (N) leads to
$$
{\rm artanh}\frac{S^{\prime}(x)}{k} 
~\left\{\begin{array}{c}\geq\cr\leq\end{array}\right\}~ k(x\!-\!x_0),\tag{O} $$
or equivalently,
$$S^{\prime}(x)
~\left\{\begin{array}{c}\geq\cr\leq\end{array}\right\}~
k\tanh k(x\!-\!x_0),\tag{P} $$
in a neighborhood of $x_1$ in the 2 cases, respectively.
However, ineq. (P) would violate ineq. (J) in the limit $x\to\pm\infty$, respectively. Contradiction. Hence ineq. (L) holds.


*Integration of ineq. (L) leads to
$$ \forall |x|\geq K\exists C>0:~~ S(x) ~\leq~-k|x|+\ln C. \tag{Q} $$
Now apply the exponential function on each side of the ineq. (Q) to derive the sought-for ineq. (D). $\Box$
--
$^1$ We allow $V(x)-E$ to be complex to include the case of unstable/decaying or resonant states.
A: I think Griffiths means that we assume that
\begin{equation} \lim_{x \rightarrow \infty} V(x) = V_R > E \end{equation}
i.e. that the potential tends to a constant value $V_R$ that is larger than $E$. In this case we can prove it as follows:
As OP writes, this means that for some $X_R$ and some $\epsilon > 0$, we have
\begin{equation} | V(x) - V_R | < \epsilon \quad \forall x > X_R \end{equation}
Let us choose $X_R$ large enough so that
\begin{equation} V(x) > E \quad \forall x>X_R \end{equation}
and approximate $V(x)$ by $V_R$ in this region. The error will be $\epsilon$, which we can make as small as we want by choosing $X_R$ large enough (but finite). 
In this case, the Schrödinger equation reads
\begin{equation} -\frac{\hbar^2}{2m}\psi_E''(x) + V_R \psi_E(x) = E\psi_E(x) \quad \forall x>X_R \end{equation}
which means
\begin{equation} \psi_E''(x) = \frac{2m}{\hbar^2} (V_R - E) \psi_E(x) \quad \forall x>X_R \end{equation}
Since $V_R - E > 0$ by assumption, this is a differential equation with solution
\begin{equation} \psi_E (x) = C_1 \exp \left(x \sqrt{\frac{2m(V_R - E)}{\hbar^2}} \right) + C_2 \left(-x \sqrt{\frac{2m(V_R - E)}{\hbar^2}} \right) \end{equation}
Unless we want something non-normalizable, we have to set $C_1 = 0$. The value of $C_2$ must be determined by the value of $\psi_E$ and $\psi_E'$ for $x<X_R$. Hence the wave function decays exponentially. The same analysis obviously holds for negative $x$, just replace $V_R$ by $V_L$, $X_R$ by $X_L$, let $x < X_L$ etc. This is just the particle in a (reeeeeeeeally wide) finite box.
That being the case, the whole wave function is normalizable because the interval $[X_L, X_R]$ is finite and the wave function decays exponentially for $x<X_L$ and $x>X_R$.
For potentials that don't have a limit when $x \rightarrow \pm \infty$, e.g. periodic potentials, I don't think we can say much. Periodic potentials are solved by Bloch's theorem and they would give non-normalizable wave functions if we did not use periodic boundary conditions to study the bulk of the lattice.
