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.

In terms of writing the logical negation, it is best to use a universal quantifier before the $x$:
$$
\forall \varepsilon >0 \exists K>0: \forall x>K,\ \left | f(x)-l \right |<\varepsilon.
$$
The logical negation of this statement is
$$
\exists \varepsilon >0 \forall K>0: \exists x>K,\ \left | f(x)-l \right |>\varepsilon.
$$
If you think about what this means, it means that you need to be able to find arbitrarily large $x$ with $|f(x)-l|$ bigger than a specified quantity. 
For your concrete example, you can fix $\varepsilon=1/2$, say. Then, given any $K$, as the values of the sine range from $-1$ to $1$ in every interval of length $2\pi$, you can certainly find $x$ with $|\sin(x)-l|>1/2$, whatever number $l$ is. 
Once you understand this, in practice what you need to do is find two sequences $\{x_k\}$, $\{y_k\}$, both going to infinity, such that $\{\sin x_k\}$ and $\{\sin y_k\}$ converge to different values. 
