I am given


And have to prove


Which is very obvious, but I have to prove it using only the definition of the limit:

$\lim_{n\to\infty}{x_n}=a$ iff for every $\alpha>0$ there exists an $N$ such that $n>N$ implies $|x_n-a|<\alpha$

Now I started with the fact that we are asked to prove that $|\frac{2+x_n^2}{x_n} - 3|$ can be made arbitrarily small. Rewriting: $|\frac{x_n^2-3x_n+2}{x_n}| = |\frac{(x_n-1)(x_n-2)}{x_n}| = |x_n-1||x_n-2||x_n^{-1}|$

Now we can (I think) say:

Let $N_\epsilon$ be such that $n>N_\epsilon$ implies $|x_n-1|<\epsilon$ for all $\epsilon>0$

Than we know that

$|x_n-1||x_n-2||x_n^{-1}| < \epsilon(\epsilon+1)|x_n^{-1}|$ for all $n>N_\epsilon$

And now I was a bit stuck, and tried this:

We know that there exists an $N_{\frac{1}{2}}$ such that $n>N_\frac{1}{2}$ implies $|x_n-1|<\frac{1}{2}$

This means $x_n > \frac{1}{2}$ for all $n>N_\frac{1}{2}$. Which in turn means $0<x_n^{-1}<2$ for all $n>N_\frac{1}{2}$

So now we know(?)

$|x_n-1||x_n-2||x_n^{-1}| < \epsilon(\epsilon+1)2$ for all $n>\max{N_\epsilon,N_{\frac{1}{2}}}$

And because $\epsilon(\epsilon+1)2$ can take on any value $>0$ we have shown that $\lim_{n\to\infty}{\frac{2+x_n^2}{x_n}}=3$

Now this seems VERY ugly and long and just not the way this is supposed to be done. Can someone please tell me A: if this proof is even correct, and B: a nicer way to prove this (remember it MUST be done using only the above definition of a limit and elementary absolute value stuff)


  • $\begingroup$ It is basically correct. First make sure $|x_n-1|\lt 1/2$. That makes the denominator $\gt 1/2$ and makes $|x_n-2|\lt 3/2$, so it makes the absolute value of the whole thing less than $3|x-1|$. Now it is enough to make $|x_n-1|\lt \epsilon/3$. So we can write: There is an $N$ such that if $n\gt N$ then $|x-1|\lt \min(1/2,\epsilon/3)$. For any such $n$ we have $\dots$. $\endgroup$ – André Nicolas Jun 26 '14 at 19:30
  • $\begingroup$ The "Let $N_\epsilon$" overloads poor $\epsilon$, which we should remember is fixed. The for all $\epsilon\gt 0$ is in the wrong place. So it may lead an uncareful reader to conclude you are making a mistake. It is best to erase that line. $\endgroup$ – André Nicolas Jun 26 '14 at 19:53
  • $\begingroup$ @AndréNicolas Yes that was also part of my problem with the proof: I confused myself by not completely being clear on what is fixed and what we can choose and is valid 'for all ...', and so on. All in all the whole proof just seems messy and unclear to me, and I'm the one who wrote it, so that can't be a good sign I think:p $\endgroup$ – user2520938 Jun 26 '14 at 20:12

Unfortunately, proofs with this definition can be fairly tedious and long-winded for all but the most simple sequences. The only way I could think of simplifying it is as follows:

As before, we have $$\left|\frac{2+x_n^2}{x_n}-3\right| = |x_n-1|\left|1-\frac{2}{x_n}\right|$$ We can find $N$ such that $|x_n-1| < \frac{\epsilon}{3}$ and $|x_n-1| < \frac{1}{2}$ for $n > N$. From there, we have that $\left|1-\frac{2}{x_n}\right| < 3$ so

$$n > N \implies |x_n-1|\left|1-\frac{2}{x_n}\right| < 3\frac{\epsilon}{3} = \epsilon$$

This isn't a particularly pretty definition for a limit: the topological formulation (while entirely equivalent) is, in my opinion, much nicer.

  • $\begingroup$ Oke so do you agree that you HAVE to at some point say $|x_n-1|<SomeSpecificNumber$ to be able to limit $|x_n^{-1}|$? Because for some reason I had this feeling that that should not be necessary, and it seemed a bit ugly to me. I expected to be able to just keep re-writing the terms until I arrived at a solution, without having to use explicit numbers, if that makes sense? $\endgroup$ – user2520938 Jun 26 '14 at 20:15
  • $\begingroup$ Yes, because otherwise $x_n^{-1}$ can grow without bound (i.e. approach zero). $\endgroup$ – Joshua Mundinger Jun 26 '14 at 20:20
  • $\begingroup$ Yea that's what I found out too:p. Thanks for the answer and the confirmation that this explicit bound on $|x_n-1|$ is indeed required. $\endgroup$ – user2520938 Jun 26 '14 at 20:23

Try rewriting $$\frac {2+x_n^2}{x_n}= \frac {2}{x_n}+ \frac{x_n^2}{x_n} $$

Let $|x_n-1|< \epsilon$. Then $ \frac {2}{1+\epsilon}<\frac {2}{x_n} < \frac {2}{1-\epsilon}$

  • $\begingroup$ I tried that but I do not see how that simplifies things significantly. You will still need to at some point say something like $|x_n-1|<\frac{1}{2}$ I think? And that was my main problem with my proof: the way I bound $x_n^{-1}$, but I do not think it can be done differently. $\endgroup$ – user2520938 Jun 26 '14 at 19:39
  • $\begingroup$ If you make $x_n$ close-enough to $1$, then $2/x_n^2$ will be as close as you want to 2. Let me see if I can write something more specific. $\endgroup$ – user99680 Jun 26 '14 at 19:48
  • $\begingroup$ I think you did the terms in that inequality the wrong way around? $|x_n-1|<\epsilon$: $1-\epsilon < x_n < 1+\epsilon$. So $\frac{1}{1-\epsilon} > \frac{1}{x_n} > \frac{1}{1+\epsilon}$. But than we only know $\frac{1}{1-\epsilon} > \frac{1}{x_n}$, but $\frac{1}{1-\epsilon}$ is not bounded on $\epsilon>0$, so we basically only have $\infty > \frac{1}{x_n}$. Please tell me if I make a mistake here $\endgroup$ – user2520938 Jun 26 '14 at 20:03
  • $\begingroup$ @user2520938: You're right, let me edit. $\endgroup$ – user99680 Jun 26 '14 at 20:04

You can use simply arithmetic of limit:

$\lim_{n \to \infty} x_n=1$, so $\lim_{n \to \infty} (x_n)^2=(\lim_{n \to \infty} x_n)^2=1 \times 1$.

Next $\lim_{n \to \infty} x_n^2+2=3$. Finally:

$\lim_{n \to \infty} \frac{x_n^2+2}{x_n}=\frac{lim_{n \to \infty} x_n^2+2}{\lim_{n \to \infty} x_n}=\frac{3}{1}=3$

  • $\begingroup$ Yes but I'm not allowed to use that. Like I said, it must only use the definition I gave and absolute value inequalities. $\endgroup$ – user2520938 Jun 26 '14 at 19:36

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