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.
 A: Hint: Limit should be unique. Find two sequences $x_n$ and $y_n$, each one going to infinity, such that $\sin x_n$ and $\sin y_n$ converge to different values.
A: 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. 
A: 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 
$$ \forall l\> \exists \epsilon > 0\> \forall K \>\exists x: x > K \land \left|f(x) - l\right|\ge  \epsilon. $$
A: 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.
A: Actually, all you need is a single $\epsilon$ to show that the negation is true, i.e., a counter example.  So pick $\epsilon\in (0,0.5)$, and note that for all $K$, there exists $x\gt K$ such that $|\sin x-l|\ge 1\gt \epsilon$.  Showing this will require two steps, one that assumes $l\in [0,1]$ and one that assumes $l\in [-1,0)$.
In practice, the most rigorous way to demonstrate this is to use sequences (as mentioned elsewhere).  For example, consider the sequence $x_n=\pi+2\pi n+{1\over n^2+1}$.  This sequence (and many others like it) will cause $\sin(x_n)$ to approach an exact limit as $n\to\infty$ (though this particular sequence may not be helpful in your proof...).
