Find $\lim_{x \to \infty} \frac{\sin{x}+\cos{x}}{x}$ 
Find 
  $$\lim_{x \to \infty} \Bigg(\frac{\sin{x}+\cos{x}}{x}\Bigg)$$

My thinking is
$$-1 < \sin{x} < 1 $$
$$-1 < \cos{x} < 1.$$
Therefore
$$-2 < \sin{x} + \cos{x} < 2$$
$$\frac{-2}{x} < \frac{\sin{x} + \cos{x}}{x} < \frac{2}{x}.$$
But is this even allowed? If so, why? Thanks

Yes that is the correct way, indeed
$$-\frac2x\le \frac{\sin{x}+\cos{x}}{x}\le \frac2x \iff 0\le \left|\frac{\sin{x}+\cos{x}}{x}\right|\le\left|\frac 2{x}\right| \to 0$$
and we conclude by squeeze theorem.
The key point here is that since $\sin{x}+\cos{x}$ is bounded therefore, according to the definition of limit, for any $\epsilon>0$ we can find $\bar x$ such that for all $x\ge \bar x$ we have $\left|\frac{\sin{x}+\cos{x}}{x}\right|<\epsilon$ and then
$$\lim_{x \to \infty} \Bigg(\frac{\sin{x}+\cos{x}}{x}\Bigg)=0$$
