= Solution
For every admissible $m,n$, the <trigonometric inequality> $\sin z\leq z$ for $z\geq0$ gives
$$
\frac4{h^2}\sin^2\frac{m\pi h}2\leq m^2\pi^2,\qquad
\frac4{h^2}\sin^2\frac{n\pi h}2\leq n^2\pi^2.
$$
Thus $\Lambda_{m,n}\leq\lambda_{m,n}$, in particular for $m=n=N-1$ and every $N\geq3$.
For these highest-index <eigenvalues>, use $h=1/N$ and $\sin((N-1)\pi/(2N))=\cos(\pi/(2N))$ to obtain
$$
\lambda_{N-1,N-1}-\Lambda_{N-1,N-1}
=2\pi^2(N-1)^2-8N^2\cos^2\frac\pi{2N}.
$$
Divide by $N^2$ and take the <limit>:
$$
\boxed{\lim_{N\to\infty}\frac{\lambda_{N-1,N-1}-\Lambda_{N-1,N-1}}{N^2}=2\pi^2-8>0.}
$$
For example, eventually the normalized difference exceeds $(2\pi^2-8)/2$, so the unnormalized difference is at least $(\pi^2-4)N^2$ and tends to $+\infty$. This does not contradict (c): there the indices stay fixed as the grid is refined; here the <eigenfunctions> oscillate increasingly quickly, with <wavelength> comparable to the mesh spacing. It illustrates the <high-frequency eigenvalue error of the discrete Laplacian>.
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