Answer
$$\lim_{t\to0}\frac{\sin(1-\cos t)}{1-\cos t}=1$$
Work Step by Step
$$\lim_{t\to0}\frac{\sin(1-\cos t)}{1-\cos t}$$
Let's take $\theta=1-\cos t$
Then as $t\to0$, $\theta\to(1-\cos0)=1-1=0$
Therefore, $$\lim_{t\to0}\frac{\sin(1-\cos t)}{1-\cos t}=\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$
Apply Theorem 7 here: $$\lim_{t\to0}\frac{\sin(1-\cos t)}{1-\cos t}=1$$
