 Solution

The singular weight requires care about the space of admissible polynomials. Since $n\ge 1$, $x^n$ has finite squared norm. A polynomial $q$ has finite error relative to $x^n$ near zero exactly when $q(0)=0$: a nonzero constant term makes the integrand asymptotic to $q(0{)}^2/x$, whose integral diverges. If $q(0)=0$, the residual is $O(x)$ and the integrand is $O(x)$.

Thus the full space ${\Pi }_{n-1}$ is not a subspace of $V$, so the projection theorem cannot be applied to it as written. Nevertheless a unique best finite-error approximation does exist inFor $n=1$, this is the zero space and the answer is $q=0$. For $n\ge 2$, $W$ is finite dimensional and its Gram matrix is positive definite, because a nonzero polynomial cannot vanish on an interval. Solving the normal equations therefore gives its unique orthogonal projection of $x^n$. All omitted polynomials have infinite error, so this also minimizes the extended error over the full polynomial set. The distinction is an invalid full-space projection premise, not nonexistence of every finite-error best approximation. As throughout weighted $L^2$, use equivalence classes for the underlying measure.