Answer
$$\lim_{\theta\to\infty}\frac{\cos\theta-1}{\theta}=0$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{\theta\to\infty}\frac{\cos\theta-1}{\theta}$$
We know that $-1\le\cos \theta\le1$
So, $$-2\le\cos\theta-1\le0$$ $$\frac{-2}{\theta}\le\frac{\cos\theta-1}{\theta}\le0$$ (because we are considering $\theta\to\infty$, we assume that $\theta\gt0$)
And, as $\theta\to\infty$, $$\lim_{\theta\to\infty}\frac{-2}{\theta}=\lim_{\theta\to\infty}0=0$$
So, according to the Sandwich Theorem, we can conclude that $$A=\lim_{\theta\to\infty}\frac{\cos\theta-1}{\theta}=0$$
