So I have this sequence in hand: $x_n=\frac{\sqrt{n}}{n+1}$, and I can intuitively see that its limit as $n\to\infty$ is 0, and I can verify it with the $\varepsilon$ definition of the limit. But I am asked to FIND this limit with only the epsilon definition of the limit, how can I do that? (I can only find the limit by manipulating the statement into $\sqrt{\lim_{n\to\infty}{\frac{1}{n+2+\frac{1}{n}}}}$)

| cite | improve this question | | | | |
  • 3
    $\begingroup$ The $\epsilon$-$N$ definition of limit is not a tool for finding the limit. $\endgroup$ – André Nicolas Nov 9 '14 at 2:49

Note that for any $\epsilon > 0$,

$$|x_n-0|=\left| \frac{\sqrt{n}}{n+1} - 0\right| < \frac{1}{\sqrt{n}} < \epsilon,$$

if $n > 1/ \epsilon^2.$

Hence $x_n \rightarrow 0$ and a limit when it exists is unique.

| cite | improve this answer | | | | |
  • $\begingroup$ So the idea is to guess a limit then check it? Is there a way to find the limit without guessing first? $\endgroup$ – galois Dec 25 '15 at 7:26
  • $\begingroup$ It isn't exactly a guess if you consider it to be a derivation based on the assumption that the limit exists. So I guess a formal proof would involve first proving that the sequence converges, which then justifies the derivation of the limit. Alternatively you could consider it a guess, which you are then using to prove that the limit both exists and is equal to 0, in one step. At least that is my understanding $\endgroup$ – Paul Oct 18 '17 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.