Prove that a function does not have a limit as $x\rightarrow \infty $ Problem statement:
Prove that the function $f(x)=\sin x$ does not have a limit as $x\rightarrow \infty $.
Progress:
I want to construct a $\varepsilon -\delta  $-proof of this so first begin by stating that the limit actually exists:
$\lim_{x\rightarrow \infty}f(x)=l$ if $\forall \varepsilon >0 \exists K>0: x>K \Rightarrow \left | f(x)-l \right |<\varepsilon$
Now we want to prove a negation(correct?), that is:
$\forall \varepsilon >0 \nexists K>0: x>K \Rightarrow \left | \sin x-l \right |<\varepsilon$. But how do I prove that there doesn't exist such a K?
I do know that $\left | \sin x -l \right |$ can be simplified(say using the triangle inequality) but I am no longer able to just find a contradiction but actually construct a direct proof.

As $|\sin x|\leq1$ for all $x$ any number $\eta$ aspiring to be $\lim_{x\to\infty}\sin x$ would have  to lie in the interval $[-1,1]$. But given  any  $\eta\in[-1,1]$ there are arbitrarily large $x$ such that $|\sin x-\eta|\geq1$. It follows that $\lim_{x\to\infty}\sin x=\eta$ is impossible.
