The Weyl law for a semiclassical Schrodinger operator 
$$ A_h\ := \ -h^2\Delta+V(x) $$
on an $d$-dimensional complete Riemannian  manifold $M$ 
says that the number $N(A_h,1)$ of eigenvalues of $A_h$ which are smaller than  1 has asymptotic behavior 
$$
	N(A_h,1)\  \sim \ 
	\frac1{(2\pi h)^d}\ \text{Vol}\
	{\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le 1{\Large)}, \quad h\to 0.\ \ \ (\ast)
$$
I am interested in a non-semiclassical Schrodinger operator 
$$ A\ := \ -\Delta+V(x).$$
I believe that the number $N(A,\lambda)$ of eigenvalues of $A$ smaller than $\lambda$ has a similar asymptotic
$$ N(A,\lambda) \sim \ \frac1{(2\pi)^d} \text{Vol}\ 
{\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le \lambda{\Large)}, \qquad h\to 0.
\quad(\ast\ast)
$$
This is, of course, true on a compact manifolds, due to the classical Weyl's law. It is also not difficult to verify (**) for operator $-\Delta+|x|^n$ on $\mathbb{R}^d$ (where $A_h$ and $A$ can be related by rescaling of $x$). 

So my question: is (**) true and, if it is true, where can I find it? 

PS. Victor Ivrii in his  review article https://arxiv.org/abs/1608.03963 mentions (page 5) that, using Birman-Schwinger principle, one can obtain a Weyl law for $N(A,\lambda)$ from (*). But I don't see how it can be done.