# 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 $x$?

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We are not interested in finding the cheapest $K$ such that $x/(1+x^2) <\epsilon$ for $x>K$. Any $K$ that works will do. It is useful to have an eye for a simple expression $E(x)$ which goes to $0$ and such that $x/(1+x^2)<E(x)$, at least for large $x$. Here it is easy, since for positive $x$, $x/(1+x^2)<x/x^2=1/x$. So if we make $1/x<\epsilon$, we will automatically have $x/(1+x^2) <\epsilon$. –  André Nicolas Oct 14 '11 at 6:40
No, it is certainly not acceptable to have $K$ in terms of $x$. –  GEdgar Oct 14 '11 at 15:38

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.

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Can you explain why 1/e is okay? I understand that mathematically for the function 1/x, but how do we arrive at it mathematically for this function? –  Jon Martin Oct 14 '11 at 5:41
If $x\geqslant1/\varepsilon$, $x/(1+x^2)<1/x\leqslant\varepsilon$. By the way, if you know that $g(x)=1/x$ is such that $g(x)\to0$, you can simply use the fact that $|f(x)|\leqslant|g(x)|$ for every $x$. –  Did Oct 14 '11 at 5:44
Whoops - I misread the question! Thanks for pointing out my mistake - I read it as depending on epsilon. –  mixedmath Oct 14 '11 at 5:47
Thank you! Makes perfect sense. –  Jon Martin Oct 14 '11 at 5:57

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.

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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 $K=K(\epsilon):={1\over\epsilon}$.

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