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

Solution 1:

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

$$\lim_{x \to \infty} \Bigg(\frac{\sin{x}+\cos{x}}{x}\Bigg)=0$$

Solution 2:

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.

Solution 3:

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.