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.

You are taking the negation of your statement wrongly. Let's start with some statement $\forall x. \phi(x)$, which tells us that "$\phi(x)$ holds for every $x$", its negation (in natural language is): "$\phi(x)$ is non allways right" or "$\phi(x)$ is sometimes wrong", as a formula: $\exists x. \neg\phi (x)$. Along the same lines we find that the negation of $\exists x. \phi(x)$ is $\forall x. \neg \phi (x)$. So negating the statement 
$$ \exists l\> \forall \epsilon>0 \> \exists K\> \forall x:  x > K \Rightarrow \left|f(x) - l\right|< \epsilon $$
we get 
$$ \forall l\> \exists \epsilon > 0\> \forall K \>\exists x: x > K \land \left|f(x) - l\right|\ge  \epsilon. $$
