Small angle increase to increment of $\frac{\sin x}{x}$

Question

Question: When $x$ increases from $\pi$ to $\pi+\epsilon$, where $\epsilon$ is small, the increment in $$\frac{\sin x}{x}$$ is approximately equal to $p\epsilon$. Find $p$ in terms of $\pi$.

My calculation: $$p\epsilon=\frac{\sin(\pi+\epsilon)}{\pi+\epsilon}-\frac{\sin\pi}{\pi}=\frac{\sin\pi \cos\epsilon+ \cos\pi \sin\epsilon}{\pi+\epsilon}-\frac{\sin\pi}{\pi}=\frac{-\sin\epsilon}{\pi+\epsilon}$$ which, for small values of $\epsilon$, $$\approx\frac{-\epsilon}{\pi+\epsilon}$$ $$\therefore p\approx-\frac{1}{\pi+\epsilon}$$ or, as $\epsilon\to0$, $$p\to-\frac{1}{\pi}$$ However, the solution provided is: $$p=-\frac{1}{\pi}$$ Am I missing something?

Answer 1

You are doing fine. The question is asking for the first order approximation.

\begin{align}-\sin \epsilon(\pi+\epsilon)^{-1}&=-\frac{\sin \epsilon}{\pi}\left(1+\frac{\epsilon}{\pi} \right)^{-1}\\ &=-\frac{1}{\pi}\left(\epsilon + O(\epsilon^2) \right)\left(1-\frac{\epsilon}{\pi}+O(\epsilon^2)\right)\\ &=-\frac1{\pi}\left(\epsilon +O(\epsilon^2) \right)\end{align}

Hence, $p$ is $-\frac1{\pi}$.