<s>[INST] How can I prove $\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$ How can I prove(using sequence convergence definition):
 $$\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$$
I need to cancel the n in the numerator, any hint will be appreciated.
 [/INST] For a given $\epsilon > 0$, we have: $|a_n-1|=\left|\dfrac{n}{\sqrt{n^2+n+1}}-1\right|= \left|\dfrac{n-\sqrt{n^2+n+1}}{\sqrt{n^2+n+1}}\right|= \dfrac{n+1}{\sqrt{n^2+n+1}\left(\sqrt{n^2+n+1}+n\right)}< \dfrac{n+1}{n(n+n)}< \dfrac{2n}{2n^2} = \dfrac{1}{n}< \epsilon$, when $n  > 1/\epsilon$. Thus you simply choose $N_0 = 1+\lfloor 1/\epsilon\rfloor$, then if $n \ge N_0$ then: $|a_n - 1| < \epsilon$ which means $a_n \to 1$. 
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