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
 A: 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
$$\lim_{x \to \infty} \Bigg(\frac{\sin{x}+\cos{x}}{x}\Bigg)=0$$
A: You are using the squeeze theorem.  It is one of the few tools for limits that does not require continuity and gives the existence of the limit for free.
Using your example, the argument (using the even lazier $|\sin x + \cos x| \leq 2$) is


*

*Since $\sin x + \cos x \leq 2$ everywhere, $\frac{\sin x + \cos x}{x} \leq \frac{2}{x}$ everywhere, so, if it exists, whatever the limit is as $x \rightarrow \infty$, it is less than or equal to the same limit of $2/x$.

*Since $ -2 \leq \sin x + \cos x$ everywhere, $\frac{-2}{x} \leq \frac{\sin x + \cos x}{x}$, so, if it exists, whatever the limit is as $x \rightarrow \infty$, it is greater than or equal to the same limit of $-2/x$.

*$\lim_{x \rightarrow \infty} \frac{2}{x} = 0$ and $\lim_{x \rightarrow \infty} \frac{-2}{x} = 0$, both of which you seem to be able to justify.  So $$  0 \leq \lim_{x \rightarrow \infty} \frac{\sin x + \cos x}{x} \leq 0  \text{,}  $$ forcing $\lim_{x \rightarrow \infty} \frac{\sin x + \cos x}{x} = 0$.


The last step is where the name comes from.  The two easy functions bound the limit above and below, squeezing together to give the limit of the complicated function nowhere to go except to the three functions' common limit.
A: And what's not allowed in it?
You're giving a lower-bound and an upper-bound for your function. Since you're looking for the limit at $x\to +\infty$, you can specify that your inequality holds for $x>0$ (your function is undefined at $0$, and the inequality should be reversed for negative $x$). So all you need is to add: let $x>0$.
Then since $\lim\limits_{x\to +\infty}\dfrac{1}{x}=0$, according to the squeeze theorem, you get your desired result.
A: Yes, that’s simply the Squeze Theorem, which states if $g(x)<f(x)<h(x)$ as $x$ approaches $a$ and $g(x) = h(x) = L$, then:  $$\lim_{x\to a} f(x) = L$$ 
$$-1\leq\sin x\leq 1$$ $$-1\leq\cos x\leq 1$$ $$\implies -2\leq\sin x+\cos x\leq 2$$ $$\implies -\frac{2}{x}\leq \frac{\sin x+\cos x}{x}\leq \frac{2}{x}$$ $$\implies \lim_{x\to \infty}-\frac{2}{x}\leq \lim_{x\to \infty}\frac{\sin x+\cos x}{x}\leq \lim_{x\to \infty}\frac{2}{x}$$ $$\implies 0\leq\lim_{x\to \infty}\frac{\sin x+\cos x}{x}\leq 0$$
Therefore, the limit of the function is $0$.  $$\lim_{x\to \infty}\Bigg(\frac{\sin x+\cos x}{x}\Bigg) = 0$$
A: The functions $\mathrm{sin} x$ and $\mathrm{cos} x$, being complementary to each other, the exact bounds would be:
$$-\sqrt2 \le \mathrm{sin}x + \mathrm{cos} x \le \sqrt2$$
$$-\frac{\sqrt2}{x} \le \frac{\mathrm{sin} x + \mathrm{cos} x}{x} \le \frac{\sqrt2}{x}$$
You may use trigonometric simplification to find the bounds.
Hint:
$$-\sqrt2 \le \sqrt2\mathrm{sin}\left(x +\frac{\pi}{4}\right)\le \sqrt2$$
And the rest you know. Anyway, you didn't have to know the exact bounds, just knowing that the numerator is finite should be enough to make the limit $0$. Just keep this thing in the back of your mind that both are complimentary, it might be helpful when you require exact bounds.
Consider this solution a result of a trigonometric simplification and as others have mentioned you may use squeeze theorem too, like you already have.
