How would you show that $\displaystyle \lim_{x \to \pi}\sin x = 0$

Question

Based on my last question (Questions about the Epsilon-Delta definition of limit and understanding what it is saying completley), I was wondering how you would show $\displaystyle \lim_{x \to \pi}\sin x = 0$. I know how to prove $\displaystyle \lim_{x \to 0}\sin x = 0$

Given $\varepsilon > 0$, Let $\varepsilon = \delta$. Suppose $0 < \mid x \mid < \delta$. Check $\mid \sin x \mid < \varepsilon$

Since $\mid \sin x \mid < \mid x \mid$,

$\mid \sin x \mid < \mid x \mid < \delta = \varepsilon$

so $\mid \sin x \mid < \varepsilon$. $\blacksquare$

But when trying to prove:

Given $\varepsilon > 0$, Let $\varepsilon = ?$. Suppose $0 < \mid x - \pi\mid < \delta$. Check $\mid \sin x - 0 \mid < \varepsilon$

I do not how to write delta in terms of epsilon because of the $\pi$. All I have gotten is that

$\mid x - \pi + \pi \mid < \delta + \pi$

so $\mid \sin x \mid < \mid x \mid < \delta + \pi$

I was thinking that maybe $\delta = \varepsilon - \pi$, so that

$\mid \sin x \mid < \mid x \mid < (\varepsilon - \pi) + \pi$

$\mid \sin x \mid < \varepsilon$

For small $\varepsilon$, $\delta$ will be negative, so maybe $\delta = \mid \varepsilon - \pi \mid$? I'm not sure...

But $\delta = \varepsilon$ should be true as well. It would make sense to me that since as $x$ approaches $0$ gets the same answer as $x$ approaches $\pi$, the delta would be the same (the function is non monotonic, so the behaviour around zero is the same as around $\pi$ I think). Also, based on this graph https://www.desmos.com/calculator/q6uu2ofasr, $\delta = \varepsilon$ should work, so I do not know how to find the answer, or if my reasoning and answer are correct. Some feedback on where I went wrong (if I did) and how to actually solve the problem would be appreciated! Also, if my proof writing is bad, some tips on how to make it more clear or 'better' (objectively) would also be appreciated!

Answer 1

Let $y=x-\pi$. Then, we see that for any $\varepsilon>0$

$$\begin{align} |\sin(x)|&=|\sin(y+\pi)|\\\\ &=|\sin(y)|\\\\ &\le |y|\\\\ &=|x-\pi|\\\\ &<\varepsilon \end{align}$$

whenever $|x-\pi|<\delta=\varepsilon$. And we are done!

Answer 2

Note that $|\sin x-\color{blue}{0}|=|\sin x-\color{blue}{\sin \pi}|=2|\cos\frac{x+\pi}{2}||\sin \frac{x-\pi}2|\leq 2|\sin\frac{x-\pi}2|\leq|x-\pi|\tag 1$

The last inequality is due to the inequality $|\sin y|\leq y$ for all $y\in \mathbb R$. The second equality is due to the identity: $\sin A-\sin B=2\cos \frac{A+B}2\sin \frac{A-B}2$.

For any $\epsilon\gt 0,$ choose $\delta=\epsilon$ and the result follows by $(1)$.