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