Limit of $\sin(\pi n)$ as $n$ approaches infinity

Question

I'm in calculus $2$ where we are learning about sequences, series, and infinite sums and I'm looking over my teacher's class notes and in them she wrote: $$ \lim_{x\to \infty} \sin(\pi x) = \text{DNE}$$ $$\lim_{n\to \infty} \sin(\pi n) = \lim_{n\to \infty} 0 = 0$$ can someone explain why this is? I'm not sure if this makes sense, it may have needed context from the class but I don't remember her going over this so I can't recall what she may have said

Answer 1

For $n\in \mathbb{Z}$, $\sin(n\pi)=0$. That is to say, the sine function is $0$ whenever its argument is an integer multiple of $\pi$.

Therefore, we have

$$\lim_{n\to \infty}\underbrace{\sin(n\pi)}_{=0}=\lim_{n\to \infty}0=0$$

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If $x\in \mathbb{R}$, then $\sin(\pi x)$ is a continuous, periodic function with period $2$. So, no matter how large we take $x$, there is a point beyond $x$ for which $\sin(\pi x)$ attains any value of its range $[-1,1]$. Therefore, the limit fails to exist.

Answer 2

This is an example of the existence of the limit of a sequence, but does not exist when we convert it into a function, then:

Limits that do not exist:

1. $$\lim_{x\to +\infty}\sin x$$
2. $$\lim_{x\to -\infty}\sin x$$
3. $$\lim_{x\to +\infty}\cos x$$
4. $$\lim_{x\to -\infty}\cos x$$
5. $$\lim_{x\to +\infty}\tan x$$
6. $$\lim_{x\to -\infty}\tan x$$