I'm trying to prove this limit here

Prove $\lim_{x \to \infty}\frac{2|x|}{x+1} = 2$, using epsilon-delta or sequence limit definition

Here's the answer I've come up so far

Let $\epsilon > 0 $, by Archimedian Property, then exist $m \in N$ such that

Consider $\epsilon ' = \frac{1}{2}\epsilon$, it's clear that $\epsilon ' > 0$ then we get $\frac{1}{m} < \epsilon '$

Then, for every $x ≥ m$, we get

$|\frac{2|x|}{x+1} - 2| = |\frac{2|x|-2x-2}{x+1}| = |\frac{2x-2x-2}{x+1} | = |\frac{-2}{x+1} | = \frac{2}{x+1} ≤ \frac {2}{x} \leq \frac{2}{m} < 2\epsilon ' = \epsilon $

limit proven.

Is this correct? also it seems there is another way to prove this? like using epsilon-delta definition.

Any insight would really help, thanks beforehand.

  • $\begingroup$ "Using definition" is quite unclear. What does that mean? $\endgroup$ – David G. Stork Oct 19 '20 at 1:20
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    $\begingroup$ Your manipulation with the inequalities seems like the right strategy. I think you need to be more careful with your $\varepsilon$. Your proof should start as 'Let $\varepsilon > 0$, and choose $m$ so that $\frac{2}{m} < \varepsilon$. Then (inequality stuff)'. $\endgroup$ – Square Oct 19 '20 at 1:32
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    $\begingroup$ You start with an arbitrary $\epsilon$ and prove that after some point, the the fraction is within $\epsilon$-neighborhood of the number 2. So I'd say your proof is correct. $\endgroup$ – PkT Oct 19 '20 at 1:51
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    $\begingroup$ "delta epsilon" is short hand for any of the four types of proof: $\delta-\epsilon$, $\delta-M$, $\epsilon-N$ or $N-M$. You use $\delta-\epsilon$ to prove $\lim_{x\to c}f(x)=L$. You use $\delta-M$ to prove $\lim_{x\to c}f(x)=\pm \infty$. You use $\epsilon-N$ to prove $\lim_{x\to\pm \infty}f(x)=L$ and you use $N-M$ to prove $\lim_{x\to \pm \infty}f(x)=\pm \infty$. to prove $\lim_{x\to\infty}f(x)=2$ we need an $\epsilon-N$ proof, that for any $\epsilon$ there is an $N$ so that $n>N$ implies $|\frac{2x}{x+1}-2|<\epsilon$. That IS the type of proof you gave and it was fine. $\endgroup$ – fleablood Oct 19 '20 at 23:17
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    $\begingroup$ " like using epsilon-delta definition." ..... so you DID you the epsilon-delta definition. You just had to use $n > M \implies....$ rather then $|x-c|< \delta$ because you have $x\to \infty$ you cant have $|x - \infty| < \delta$. What you choose instead is $x > N$. $\endgroup$ – fleablood Oct 19 '20 at 23:19

Although the OP's intent can be understood, it is advised that they carefully "'dot their i's" and "cross their t's" to logically nail it down and they can then compare their technique to the one given here.

The OP should understand that

When taking a limit out to infinity, such as

$\quad \displaystyle \lim_{x \to +\infty} f(x) = L$

that the $\delta \gt 0$ 'bit' has a different interpretation (see the last section).

Set $f(x) = \frac{2|x|}{x+1}$. It is easily shown (use inequality algebra) that

$\quad \displaystyle f\bigr(\,[0,+\infty)\,\bigr) \subset [0, 2]$

So we now only have to address a challenge $\varepsilon$ satisfying $0 \lt \varepsilon \lt 2$.

For $x \gt 0$

$\quad f(x) \ge 2-\varepsilon \text{ iff }$
$\quad \quad 2x \ge 2x + 2 -\varepsilon x - \varepsilon \text{ iff }$
$\quad \quad \varepsilon x \ge 2 - \varepsilon \text{ iff }$
$\quad \quad x \ge \frac{2 - \varepsilon}{\varepsilon}$

Setting $d = \frac{2 - \varepsilon}{\varepsilon}$ we can now write as true

$\quad \displaystyle f\bigr(\,[d,+\infty)\,\bigr) \subset [2 - \varepsilon, 2 + \varepsilon]$

and so

$\quad \displaystyle \lim_{x \to \infty}\frac{2|x|}{x+1} = 2$

The reader can review the definition

$\quad$ Limits at infinity

This definition uses strict inequalities and the limit control variable is designated with the letter $c$, but the above is an equivalent formulation.

Heck, you could even use $\delta$ rather that $c$ or $d$, but that would bring a frown to the face of some mathematicians.


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