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.
Write $n^{1/n} = 1 + a_n.$ We have $0&lt;a_n$ for all $n.$ The binomial formula shows

$$n = (1 + a_n)^n = 1 + na_n + [n(n-1)/2]a_n^2 + \cdots$$

for $n&gt;1.$ Thus for $n&gt;1$ we have $n&gt;[n(n-1)/2]a_n^2,$ which implies $0 &lt; a_n &lt; \sqrt {2/(n-1)}.$ It follows that $a_n \to 0,$ giving $n^{1/n} \to 1.$
see my nice answer:
$$\lim_{n\to \infty}n^{1/n}=e^{\lim_{n\to \infty}\large \frac1n\ln n}$$ 
apply l'hopital's rule,
$$=e^{\lim_{n\to \infty}\large \frac{1/n}{1}}=e^{\lim_{n\to \infty}\large \frac1n}=1$$