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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 $\sqrt{\lim_{n\to\infty}{\frac{1}{n+2+\frac{1}{n}}}}$)

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

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  • $\begingroup$ 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

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