How to approach proving the limit of $[\cos(x) - 1]$ doesn't exist? I'm wondering what the most graceful/accurate way to state the limit of $\cos(x) - 1$ as $x$ approaches $-\infty$ doesn't exist. Intuitively, this limit can't exist; but I'm unsure if using the fact that all subsequences of a sequence must converge for the limit to exist would work here (My original approach) as this seems more like the limit of a function rather than a sequence (Even when you define a sequence as having elements equivalent to this).
Could I get some opinions/headings?
Hint:

Check the sequences $${x_n}=-\frac{2n+1}{2}\pi$$ and $${y_n}=-4n\pi$$, converging to negative infinity. Then put them in the function and find it violates seqential definition of limit of function.
A periodic function can't have a limit at infinity.
Because a periodic function is not constant and in a period you will find a value that differs from the would-be limit by a finite amount. This difference repeats in every period.

Formally:
$$\forall\epsilon&gt;0:\exists M:\forall x&gt;M\implies |p(x)-L|&lt;\epsilon.$$
Let us take any $x_0$ such that $|p(x_0)-L|=\eta&gt;0$; such an $x_0$ necessarily exists as the function is non-constant. Then by periodicity,
$$\forall\epsilon&gt;0:\exists M: x_0+kT&gt;M\implies |p(x_0+kT)-L|=\eta&lt;\epsilon$$ is false for $\epsilon&lt;\eta$.