How can I prove $\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$

Question

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.

Answer 1

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 $a_n \to 1$.

Answer 2

HINT

$$ \frac{n}{\sqrt{n^2+n+1}} = \sqrt{\frac{n^2}{n^2+n+1}} = \sqrt{\frac{1}{1+\frac{1}{n}+\frac{1}{n^2}}} $$

Answer 3

$\dfrac{n}{\sqrt{(n+1)^2}} =:a_n\lt $

$\dfrac{n}{\sqrt{n^2+n+1}} =: c_n \lt \dfrac{n}{\sqrt{n^2}} =:b_n.$

$a_n =\dfrac{n}{n+1} \lt c_n \lt \dfrac{n}{n} = 1 = b_n.$

$\lim_{n \rightarrow \infty} a_n \le c_n \le b_n =1.$

Left to show: $\lim_{n \rightarrow \infty} a_n =1.$

Can you do it?

Note: $a_n= 1- \dfrac{1}{n+1}.$

Answer 4

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,$$\lim\limits_{x\to\infty}\frac x{\sqrt{1+x+x^2}}=\color{blue}{\lim\limits_{x\to\infty}\frac 1{\sqrt{\frac 1{x^2}+\frac 1x+1}}=1}$$