Consider the one dimensional Schrödinger hamiltonian $\mathcal{H}=-\frac{\hbar^2}{2} \frac{d^2}{dx^2} + V(x)$.


Suppose that $V:\mathbb{R} \rightarrow \mathbb{R}^+$ is a continuous and confining potential $\displaystyle \lim\_{\lvert x\rvert \to +\infty} V(x)=+\infty$


It is well known that $\mathcal{H}$ has pure discrete spectrum $\lambda\_1< \cdots \leq \lambda\_n$ with $\lambda\_n \to +\infty$ as $n\to + \infty$. 


Furthermore, if $V$ is a polynomial of even degree $2s$, then the eigenvalues obey the asymptotic formula


$$\lambda \_k \sim {C\_{s,\hbar}\, k}^{\frac{2s}{s+1}}$$


**I want to know if the above asymptotic formula holds if**, instead of having a polynomial potential as above, I consider a "pseudo-polynomial" in the sense that $V$ is another function on a segment $[a,b]\subset \mathbb{R}$ (like $\sin(x)$) and outside $\mathbb{R}\setminus[a,b]$ its an even polynomial such that the result is a continuous piecewise function. 



Pseudo-polynomial potentials for Schrödinger operators
The answer is yes. Suppose your $V$ equals to a polynomial $P$ when $|x|$ is large. Then there is a constant $c$ such that we have
$P-c<V<P+c$, which implies that $\lambda\_k^\prime-c<\lambda\_k<\lambda^\prime\_k+c$,
where $\lambda\_k^\prime$ are the eigenvalues of the potential $P$.
So $\lambda\_k$ and $\lambda\_k^\prime$ have the same asymptotics.


EDIT. I suppose it is clear (from physical interpretation, or from the analogy with matrices) that bigger potential gives bigger eigenvalues, but for a formal proof of this one can refer on Sturm's comparison theorem.


