<s>[INST] Show that $\lim\limits_{n\to\infty}\sqrt [n] {n^{2}+n+1}=1$ Proof. Note that $1\leq \sqrt [n] {n^2+n+1}\leq \sqrt [n] {3n^{2}}$. But,
$\lim\limits_{n\to\infty}\sqrt [n] {3n^{2}} =1$. Therefore, by the squeeze lemma, we have $\lim\limits_{n\to\infty} \sqrt [n]{n^2+n+1}  =1$
My question is: How did $\lim\limits_{n\to\infty} \sqrt [n] {3n^2} =1$ be? So, how can it prove as primitive methods? We cannot use L'hopital rule, rules of derivative, exp functions.
 [/INST] We can write $\sqrt[n]{3n^2} = (\sqrt{3}n)^\frac{2}{n} = e^\frac{2\ln(\sqrt{3}n)}{n}$. Note that for $n > 1$ we have $0 < \ln(n) < n$. If $n > 1$ then $\sqrt{\sqrt{3}n} > 1$ (since $\sqrt{3}$ > 1). Thus we have for $n > 1$
\begin{align}
0 < \ln(\sqrt{\sqrt{3}n}) < \sqrt{\sqrt{3}n}\\
0 < 2\ln(\sqrt{\sqrt{3}n}) < 2\sqrt{\sqrt{3}n}\\
0 < \ln(\sqrt{3}n) < 2\sqrt{\sqrt{3}n}\\
0 < \frac{\ln(\sqrt{3}n)}{n} <\frac{2\sqrt{\sqrt{3}} \sqrt{n}}{\sqrt{n} \sqrt{n}} = \frac{2\sqrt[4]{3}}{\sqrt{n}}
\end{align} 
Since $\lim_{n \to \infty} \frac{2\sqrt[4]{3}}{\sqrt{n}} = 0$ we have by the squeeze theorem $\lim_{n \to \infty} \frac{\ln(\sqrt{3}n)}{n} = 0$. Thus we have
\begin{equation}
\lim_{n\to\infty} \sqrt[n]{3n^2} = \lim_{n \to \infty} e^\frac{2\ln(\sqrt{3}n)}{n} = e^{2\lim_{n\to\infty} \frac{\ln(\sqrt{3}n)}{n}} = e^{2(0)} = e^0 = 1
\end{equation}
</s>