Question

$\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-[\sqrt{n^2+n+1}\left]\right)$ where [ ] denotes the greatest integer function is

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

Answer 1

Answer: (B)

$\frac{1}{2}$

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

$\Rightarrow \left[\sqrt{n^2+n+1}\right]=n$

$L=\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-n)$
$=\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-n)\frac{(\sqrt{n^2+n+1}+n)}{(\sqrt{n^2+n+1}+n)}$
$=\underset{n\to \infty }{lim}\frac{n+1}{(\sqrt{n^2+n+1}+n)}$
$=\frac{1}{2}$