= Solution
In terms of $R=\ell_2/\ell_1$,
$$
\lambda_\pm(R)=\frac{\omega_\pm^2}{\omega_0^2}
=\frac12+\frac1R\pm\sqrt{\frac14+\frac1{R^2}}.
$$
Both curves decrease monotonically: with $q=1/R$, $d\lambda_\pm/dq=1\pm q/\sqrt{1/4+q^2}>0$, while $dq/dR<0$. The limits, including their leading approach, are
$$
\boxed{\begin{aligned}
R\to0^+:&\quad\lambda_- =\frac12-\frac R8+O(R^3)\to\frac12,\\
&\quad\lambda_+=\frac2R+\frac12+O(R)\to\infty;\\
R\to\infty:&\quad\lambda_-=\frac1R-\frac1{R^2}+O(R^{-4})\to0,\\
&\quad\lambda_+=1+\frac1R+O(R^{-2})\to1.
\end{aligned}}
$$
Thus the limiting squared <angular frequencies> are $\omega_0^2/2$ and infinity for short joining fibre, and zero and $\omega_0^2$ for very long joining fibre. For orientation, at $R=1$ the values are $(3\pm\sqrt5)\omega_0^2/2$. The high-frequency curve stays above $\omega_0^2$ and the low-frequency curve below $\omega_0^2/2$.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-1-b7-torsional-modes.png]
{title=Calculated squared normal-mode frequencies versus fibre-length ratio; both axes are logarithmic to show all four asymptotic limits}
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