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I've been asked to prove using definition of limits that limit of sequence $\dfrac{n}{n^2+1}$ is $0$. But when proceeding as usual I am not being able to express $n$ in terms of $\varepsilon$.

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  • $\begingroup$ Given $\epsilon > 0$, fix any integer $N$ satisfying $N > 1/\epsilon$. Then for all $n > N$ we have $$0 < \frac{n}{n^2+1} < \frac{1}{n} < \frac{1}{N} < \epsilon$$ $\endgroup$
    – user169852
    Oct 26, 2017 at 13:56

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suppose $n \over n^2+1$ as $n$ goes to infinity is 0 so for any $\epsilon>0$ there exist $N$ such that for any $n>N$ then $| {n \over n^2+1}-0|< \epsilon$ ,so we need proof there is always an $N$ exist,we simplified absolute value so we get $| {n \over n^2+1}|< \epsilon$ because $n$ is always positive we get ${n \over n^2+1}< \epsilon $ it is easy to see that ${n \over n^2+1} < {{n \over n^2}= {1 \over n}}$ so if ${ 1 \over n} < \epsilon$ the inequality is always true and finally we only need to set $N={1 \over \epsilon } <n$.

so for any $n$ with this condition the inequality i true , thus the limit of sequence is zero.

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