Analysis question limit at infinity $\lim_{n \to \infty} (-1)^n \cos(n)$

Question

1_I have questions about that when is it necessary to use bounding for limit problems e.g. this limit: $$\lim_{x\to 0} x \sin\left(\frac{1}{x}\right)$$ solution is zero because $|\sin\left(\frac{1}{x}\right)|\leq 1$ times zero becomes zero. But I wonder if I have this limit instead $$\lim_{n \to \infty} (-1)^n \cos(n)$$ can I bound it like this $|\cos(n)|\leq 1$ and $|(-1^n)|=1$ and the result becomes one?

2_When should someone use bounding rule when someone evaluate a limit problem?

Answer 1

You cannot bound $|{(-1)}^n|=1$. You can't take the absolute value like you can with $\sin{x}$ or $\cos{x}$. The limit does not exist because $\lim_{n \to \infty} {(-1)}^n$ could be $1$ or $-1$. Since $ \lim_{x \to \infty} \cos{x} \neq 0$, the limit could take on any value from $[-1,1]$ therefore it does not exist.

Answer 2

Your first example works and you can conclude that

$$\left|x \sin\left(\frac{1}{x}\right)\right|\le|x|\to0.$$

However, the same approach doesn't work in the second problem. While it is true that $|\cos(n)|\le 1$, it turns out that this limit doesn't exist because $(-1)^n$ alternates between $1$ and $-1$. Note $(-1)^n$ is $1$ for even $n$ and $-1$ for odd $n$. So $(-1)^n\cos(n)$ is $\cos(n)$ for even $n$ and $-\cos(n)$ for odd $n$. Therefore, if $a_n = (-1)^n\cos(n)$, then $a_{2n} = \cos(n)$ and $a_{2n+1} = -\cos(n)$. Since the subsequences $a_{2n}$ and $a_{2n+1}$ have different limits ($\lim a_{2n} = \cos(n)$ and $\lim a_{2n+1} = -\cos(n)$), the limit $\lim a_n$ does not exist.