I'm currently studying Lagrangian Mechanics, and I came across a problem that requires some approximation techniques to get an easily-evaluable differential equation. However, I am unclear on one portion of the differential equation, which is attached in the picture. Expanding out the first equation in terms of $\epsilon$ and applying a small angle approximation for $\epsilon$, we get:
$\epsilon''(l-R\epsilon-R\theta_0)-R\epsilon'^2-g\cos(\epsilon)\cos(\theta_0)+g\sin(\epsilon)\sin(\theta_0)=0\\ \epsilon''(l-R\epsilon-R\theta_0)-R\epsilon'^2-g\cos(\theta_0)+g\epsilon\sin(\theta_0)=0\\ \epsilon''(l-R\theta_0)+(-R\epsilon'^2-R\epsilon''\epsilon)+g\epsilon\sin(\theta_0)=g\cos(\theta_0)\\ \epsilon''+(\frac{-R\epsilon'^2-R\epsilon''\epsilon}{l-R\theta_0})+\frac{g\sin(\theta_0)}{l-R\theta_0}\epsilon=\frac{g\cos(\theta_0)}{l-R\theta_0}$
However, the answer key's answer implies that $\frac{-R\epsilon'^2-R\epsilon''\epsilon}{l-R\theta_0}=0$, and I am not sure why that would be true. I would assume it would be a consequence of the small angle approximation, but I am not sure. Why is the term in parentheses 0? Thank you for any help you can provide!
