Why $E$ is neglected at large and small $r$ of quantum harmonic oscillator? In obtaining radial solution of quantum oscillator why E is neglected?
Radial equation:


Resource: nouredine zettili.
 A: One does this because the original differential equation is overly difficult to solve without first examining the boundary conditions.
The point here is that you are initially only approximately solving the TISE in the regimes of small $r$ or large $r$.  This is where the boundary conditions are, so you want to get the correct behaviours at these boundaries.  For finite energy, the dominant term is, in one case, the $1/r^2$ term (assuming $\ell\ne 0$), and in the other case the $r^2$ term.
Once you have the correct behaviour near $r=0$ and at $r\to\infty$, the strategy is then to interpolate between small $r$ and large $r$ using a polynomial.
Thus you start with small $r$ (or Eq.(6.86)), and then go to large $r$ (or Eq.(6.87)) and this gives you a solution of the form
$$
U(r)\sim r^{\ell+1} e^{-\lambda r^2}f(r)\, , \tag{1}
$$
for some unknown polynomial $f$.  This way, you are guaranteed that the $U(r)$ you find will satisfy the correct boundary conditions.
You then plug (1) into the full TISE to get a differential equation for the unknown function $f$, which can be solved (after some gymnastics) by identification with a standard differential equation.
This is similar to the 1d harmonic oscillator, except there is no condition the solution for small $r$: in 3d you need this solution to go to $0$ smoothly since the system cannot have “negative values of $r$”.  In the 1d case, there is a condition at $r\to\infty$ and one often starts there to get the right behaviour, then interpolate back to finite $r$s.
A: $E$ is neglected because in both cases (where $r \rightarrow 0$ and when $r \rightarrow \infty$) there is a term multiplying $U(r)$ that gets very large when compared to the constant $E$.
When $r \rightarrow 0$, this is the term which depends on $\frac{1}{r^2}$; when $r \rightarrow \infty$ this is the term which depends on $r^2$.
