Showing error energy goes to zero
HINT: The error energy can be written as
$$\begin{align}\sum_k\big|d[k]\big|^2&amp;=\sum_k\left|x[k]-\frac{1}{2\pi}\int_{-W}^W\sum_nx[n]e^{-jn\omega}e^{jk\omega}d\omega\right|^2\\&amp;=\sum_k\left|x[k]-\sum_nx[n]\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\right|^2\tag{1}\end{align}$$
Now compute the integral
$$I(k-n,W)=\frac{1}{2\pi}\int_{-W}^We^{j(k-n)\omega}d\omega\tag{2}$$
and show that $\lim_{W\to\pi}I(k-n,W)=\delta[k-n]$. This reduces the second sum in $(1)$ to a single element and the result follows.