Prove by delta/epsilon that $\lim_{x \to \infty}{\frac{x}{1+x^2}} = 0$ Here's what we tried:
For every $\epsilon > 0$ there is a large number $K$ such that $|f(x)| < \epsilon$ when $x>K$.
Knowing that $K$ is a large positive number, take the positive absolute value of $f(x)$:
$$\displaystyle \frac{x}{1+x^2} < \epsilon$$
Solve for
$$x > \displaystyle \sqrt{\frac{x - \epsilon}{\epsilon}}$$
And thus $$K = \displaystyle \sqrt{\frac{x - \epsilon}{\epsilon}}$$.
Is it acceptable to have $K$ in terms of $x$?
 A: No it is not. Recall the definition of $f$ having limit $0$ at $+\infty$:
$$
\forall\varepsilon>0,\quad\exists K_\varepsilon\in\mathbb R,\quad\forall x\in\mathbb R,\quad x\geqslant K_\varepsilon\implies |f(x)|\leqslant\varepsilon.
$$
Since $\varepsilon$ is the only object mentioned before the introduction of $K_\varepsilon$ by $[\exists K_\varepsilon\in\mathbb R]$, $K_\varepsilon$ depends on $\varepsilon$ and on $\varepsilon$ only.
In your case, $K_\varepsilon=1/\varepsilon$ is fine.
A: If you are presented with an expression of the the form $\frac{x}{1+x^2}$ and have to compute the limit as $x$ goes to infinity, you of course try something like direct substitution first, but that gives $\frac{\infty}{\infty}$ in this case.
You would very much prefer to have no $x$ either in the denominator or in the enumerator of the fraction.  So, just divide both by $x$.
This gives $\frac{1}{\frac{1}{x}+x}$.
Now direct substitution yields the limit $\frac{1}{0+\infty}=0$.
(I know this looks sloppy, but it can be carried out rigorously.)
But you are looking for an $\epsilon$-$\delta$-argument.
At this point you realize that the $\frac{1}{x}$ in the denominator of the fraction $\frac{1}{\frac{1}{x}+x}$ only makes the value smaller.
I.e., as long as $x>0$, $\frac{1}{\frac{1}{x}+x}<\frac{1}{x}$.
This is the point where you come up with $K_{\epsilon}=\frac{1}{\epsilon}$ 
as Didier Piau did in his answer.
A: You say it yourself: You have to prove that "for every $\epsilon > 0$ there is a large number $K$ such that $|f(x)| < \epsilon$ when $x>K$".
Such a $K$ is not uniquely determined, but the admissible $K$'s will depend on $\epsilon$: The smaller your "tolerance" $\epsilon$, the farther out you will have to go to guarantee $|f(x)|<\epsilon\>$. In finding such a $K$ we can be gratuitous, since we are not required to produce the "optimal" $K$ for a given $\epsilon$. In problems of this sort it pays to replace the given $f$ with a simpler function $g$ that is of the same order of magnitude. In fact one has
$$|f(x)|={|x|\over 1+x^2}< {x\over x^2}={1\over x}\qquad (x>0),$$
so that it is enough to guarantee $g(x):={1\over x}<\epsilon$. But this is true as soon as $x>{1\over\epsilon}$. Therefore we may take $K=K(\epsilon):={1\over\epsilon}$. 
