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.

Note that$$\lim\limits_{x\to\pm\infty}\left(\frac 1{x}\right)^n=0$$With simple limits to infinity type problems, we can divide both the numerator and denominator by the greatest power and take the limit as each term tends towards infinity$$\begin{align*}\lim\limits_{x\to\infty}\frac x{\sqrt{1+x+x^2}} & =\lim\limits_{x\to\infty}\frac {x}{\sqrt{1+x+x^2}}\frac {\tfrac 1x}{\tfrac 1x}\\ & =\lim\limits_{x\to\infty}\frac 1{\sqrt{\frac 1{x^2}+\frac 1x+1}}\end{align*}$$The limit of the denominator is simple one, so therefore,$$\lim\limits_{x\to\infty}\frac x{\sqrt{1+x+x^2}}=\color{blue}{\lim\limits_{x\to\infty}\frac 1{\sqrt{\frac 1{x^2}+\frac 1x+1}}=1}$$
