= Solution
Set $z=\pi h/2$, so $0<z\leq\pi/4$. The formulas in (a) and (b) give
$$
\lambda_{1,1}-\Lambda_{1,1}=\frac8{h^2}(z^2-\sin^2z).
$$
For $0\leq t\leq\pi/4$, $0\leq\sin t\leq t$ and $1-\cos t=2\sin^2(t/2)\leq t^2/2$. Therefore
$$
0\leq z-\sin z=\int_0^z(1-\cos t)dt\leq\frac{z^3}6.
$$
These <trigonometric inequalities> imply
$$
0\leq z^2-\sin^2z=(z-\sin z)(z+\sin z)\leq\frac{z^3}6(2z)=\frac{z^4}3.
$$
Substitution yields the explicit <eigenvalue> error bound
$$
\boxed{|\lambda_{1,1}-\Lambda_{1,1}|\leq\frac{\pi^4}{6}h^2,\qquad C=\frac{\pi^4}{6}.}
$$
This constant is asymptotically sharp: <Taylor's theorem> gives $\sin^2z=z^2-z^4/3+O(z^6)$, so $(\lambda_{1,1}-\Lambda_{1,1})/h^2\to\pi^4/6$. Here $O$ denotes <Big O notation>.
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