Question

The value of $\underset{n\to \infty }{lim}(\sqrt{n^2+n+1}-\left[\sqrt{n^2+n+1}\right])$ is?

where $[\cdot ]$ denotes the greatest integer function and $n\in I$

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

Answer 1

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

Lets assume $f\left(x\right)=\sqrt{n^2+n+1}-\left[\sqrt{n^2+n+1}\right]$,
it can also be written as,
$f\left(x\right)=\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}-\left[\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}\right]$
Now,
${lim}_{n\to \infty }f\left(x\right)={lim}_{n\to \infty }\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}-\left[\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}\right]$
As $n\to \infty ,({(n+\frac{1}{2})}^2+\frac{3}{4})\to {(n+\frac{1}{2})}^2$
$\Rightarrow {lim}_{n\to \infty }\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}-\left[\sqrt{{(n+\frac{1}{2})}^2+\frac{3}{4}}\right]={lim}_{n\to \infty }\sqrt{{(n+\frac{1}{2})}^2}-\left[\sqrt{{(n+\frac{1}{2})}^2}\right]$
$\Rightarrow {lim}_{n\to \infty }(n+\frac{1}{2}-[n+\frac{1}{2}])$
$\because n$ is an integer,
$\therefore [n+\frac{1}{2}]=n$, where $[\cdot ]$ denotes the greatest integer function.
$\Rightarrow {lim}_{n\to \infty }(n+\frac{1}{2}-[n+\frac{1}{2}])=\frac{1}{2}$
Hence option B.