Please refrain from using logic symbols, as I do not understand those.

So, this is the question:

$$\lim_{x \to \infty} x = \infty$$

Proving this using the actual formal definition of a limit.

So for $x > N$ where $N$ is a natural number. we have to prove that $f(x) = x$ grows without bound.

How so?

The limit means for

$x > N$ we have $x > \epsilon$

Let $ M> N $ so then:

$f(x)$ does not have to reach certain bounds.

$M > N$ ensures that there exists $f(m) > f(N)$

So, $f(X) < \epsilon$ does not have to be true

If this it?

  • 4
    $\begingroup$ When $x$ goes to $\infty $, $x$ goes to $\infty$. $\endgroup$ – mesel Dec 14 '14 at 15:26
  • $\begingroup$ @mesel Haha, exactly my thoughts! :D $\endgroup$ – Swapnil Tripathi Dec 14 '14 at 15:43
  • $\begingroup$ @SwapnilTripathi: ${}{}$${}{}$${}{}$ :D $\endgroup$ – mesel Dec 14 '14 at 19:26

You should say at the beginning of your proof what is $\epsilon$, and use the definition of the limit. "$f(x)$ does not ahev to reach a certain bound" is not very rigorous.

I think it will be easier if you do it directly.

Let $\epsilon >0$, then for every $x> N:=\epsilon$ we have $f(x)=x > N = \epsilon$ and so $\lim_{x\to \infty}f(x)=\infty$ (by definition of the limit).

  • $\begingroup$ how can you let $N = \epsilon$? $\endgroup$ – Amad27 Dec 18 '14 at 14:15
  • $\begingroup$ By definition, $\lim_{x\to \infty}f(x)=\infty $ if for every $\epsilon>0$ there exists $N\in\Bbb R$ such that $f(x)>\epsilon$ for every $x >N$. So the structure has to be: 1) You consider $\epsilon >0$ 2) You find an $N$ so that the property is verified 3) You conclude that $\lim_{x\to \infty}f(x)=\infty$ In this case we have $f(x)=x$ so $f(x)=x>\epsilon$ if $N=\epsilon$. But you can take $\tilde{N} = 12313\epsilon + \pi^8$ if you prefer. $\endgroup$ – Surb Dec 18 '14 at 14:28
  • $\begingroup$ So we just have to prove there is a single $N$, which in this case was $N = \epsilon$? $\endgroup$ – Amad27 Dec 18 '14 at 14:33
  • $\begingroup$ Yes you just have to show the existence of one $N$ for which this is true. And since $f(x)>\epsilon$ for every $x>N$, it is also true for every $x>M$ with $M>N$... $\endgroup$ – Surb Dec 18 '14 at 14:37

I think you may be doing some of this the wrong way round. And "$f(x)$ grows without bound" is not at all formal.

Formally, for showing something tends to infinity you would need to show that for every $M$ there is some $N$ such that whenever $x > N$ we have $f(x) > M$.

[In symbols, $ \forall N \exists M$ such that $\forall x > N$ we have $f(x) > M$. Note that $\forall$ means "for all" and $\exists$ means "there exists" - these symbols are worth learning.]

To proof would then go as follows [with notes in square brackets];

Let $M \in \mathbb{R}$. [$M$ any real number]

Whenever $x > M$, we have that $x > M$. [With the notation above, $N = M$ and $f(x) = x$.]

Hence $x \rightarrow \infty$ as $x \rightarrow \infty$. [i.e. the proof is finished.]


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