Solution
= Solution
Writing $x=\cos\theta$ and taking the <limit> as $\theta\to0$ gives
$$
T_n'(1)=\lim_{\theta\to0}
\frac{n\sin(n\theta)}{\sin\theta}=n^2.
$$
Part (d), with $r=1$, shows that $T_n'$ is increasing on $[1,\infty)$, so
$$
T_n(x)-T_n(1)=\int_1^xT_n'(t)\,dt
\geq n^2(x-1).
$$
Since $T_n(1)=1$ and $n^2\geq n$,
$$
\boxed{T_n(x)\geq1+n(x-1)\quad(x\geq1)}.
$$