= Solution
{c}
We verify that the augmented <partition of an interval> in part <7/b/i> has vanishing <mesh of a partition>. Define $\phi_n(t)=t^2/(t^2+n)$ for $t\geq0$. Its <derivative> is
$$
\phi_n'(t)=\frac{2tn}{(t^2+n)^2}
=\frac1{\sqrt n}\frac{2u}{(1+u^2)^2},\qquad u=t/\sqrt n.
$$
Since $2u\leq1+u^2$ for $u\geq0$, we have $0\leq\phi_n'(t)\leq1/\sqrt n$. The <mean value theorem> therefore gives
$$
0<x_{k+1,n}-x_{k,n}\leq\frac1{\sqrt n}.
$$
The appended <interval> has length $1/(n+1)$, so $\|\mathcal P_n\|\leq\max\{n^{-1/2},(n+1)^{-1}\}\to0$ for the augmented <partition of an interval>.
The <sine> is <continuous>, hence <Riemann integrable> on $[0,1]$. By the <Riemann sum convergence theorem>, $T_n\to\int_0^1\sin t\,dt=1-\cos1$, using the <fundamental theorem of calculus> and the <cosine> <antiderivative>. As $T_n-S_n=\sin1/(n+1)\to0$, the answer is
$$
\boxed{\lim_{n\to\infty}S_n=1-\cos1.}
$$
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