Why is it that the SHO/SHM solution $$u(x)=H(x)\,e^{-x^2/2},\quad \Psi(x,t)=u(x)\,e^{-iEt/\hbar}$$ with $$H_n (x)=\sum_k^na_kx^k \tag{Hermite Polynomials}$$ is valid even when $x$ (displacement from equilibrium) becomes very large, in which case $H_n(x) \propto x^n$?
Isn't the Taylor expansion of the potential $$V(x_0+\delta x)\approx V(x_0)+\left.\frac{1}{2}\frac{d^2V}{dx^2}\right|_{x_0}(\delta x)^2$$ only valid for when $\delta x$ is small?
There are no approximations made in this solution.
The potential is exactly a quadratic. So, in the simple quantum harmonic oscillator system, you don't need to do the expansion of the potential you mention at the end of your question; it simply is exactly a quadratic.
There can be some confusion because people often say things like "this dip in X funny-looking potential is quadratic for small deviations from the middle of the dip," but the QHO is in an exactly quadratic potential, so it looks quadratic arbitrarily far from the middle.
