What is
$$\lim_{x\to\infty} \sin x$$
?
I always thought it was undefined; however, Wolfram|Alpha says that
$$\lim_{x\to\infty} \sin x = -1 \text{ to } 1$$
Now, what is the correct answer? What does it even mean if the limit is not a single, distinct value, but rather an interval?
Other than that, am I right in assuming that
$$\liminf_{x\to\infty} \sin x = -1$$
and
$$\limsup_{x\to\infty} \sin x = 1$$ ?
On
The limit doesn't exist and it can be proved formally by 2 subsequences with different limits, that is
$x_n=2n\pi+\frac{\pi}2 \to \infty \implies \sin(x_n)=1$
$x_n=2n\pi+\frac{3\pi}2 \to \infty \implies \sin(x_n)=-1$
Yes what is true is that $\liminf=-1$ and $\limsup=1$ indeed
$$-1\le\sin x \le 1$$
and we have found 2 subsequences which tends to those limits.
Take note that the sine function takes values between $-1$ and $1$. When you are talking about a limit to $\infty$, it's an undetermined number, which is infinitely large. Now, taking into account the periodicity of the sine function, there is no possible way to determine a specific value, as it entirely depends on the nature of the "infinite" number.
More specifically, $-1 \leq \sin(x) \leq 1, \; \; \forall x \in \mathbb R$. This means that for any given $x$ over the real numbers, the sine function is bounded. Thus, all you can say about an undetermined infinite limit (it does not exist talking strictly mathematics), is :
$$-1 \leq \lim_{x \to \infty} \sin(x) \leq 1$$
What you mentioned though is indeed true :
$$\liminf_{x\to\infty} \sin x = -1$$
$$\limsup_{x\to\infty} \sin x = 1$$