Question

The value of $\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-\left[\sqrt{n^2+n+1}\right]);n\in Z,$ where $[.]$ denotes the greatest integer function is

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

We know that,
$n<\sqrt{n^2+n+1}<n+1$ for large value of $n.$
Hence, $\left[\sqrt{n^2+n+1}\right]=n(\because n\in Z)$

$\therefore L=\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-n)$
$=\underset{n\to \infty }{lim}\frac{n+1}{(\sqrt{n^2+n+1}+n)}$
$=\underset{n\to \infty }{lim}\frac{1+\frac{1}{n}}{(\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1)}$
$=\frac{1}{2}(\because \text{we know, as}n\to \infty ,\frac{1}{n}\to 0)$

Alternate Solution:
Let $L=\underset{n\to \infty }{lim}\sqrt{n^2+n+1}$
$\Rightarrow L-n=\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-n)$
$=\underset{n\to \infty }{lim}\frac{n+1}{(\sqrt{n^2+n+1}+n)}$
$=\underset{n\to \infty }{lim}\frac{1+\frac{1}{n}}{(\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1)}$
$=\frac{1}{2}$
$\therefore L=\frac{1}{2}+n$
Taking greatest integer function both the sides, we get
$\therefore \left[L\right]=[\frac{1}{2}+n]=n(\because n\in Z)$

$\therefore \underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-\left[\sqrt{n^2+n+1}\right])$
$=\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-n)$
$=\frac{1}{2}$