Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$ 
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$ 
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks

I think this problem is much more easily solved in slightly greater generality. If $f$ is a continuous function such that $\lim_{x \rightarrow \infty} f(x) = \infty$ then as $x \rightarrow \infty$, for arbitrarily large $x$, $f(x)$ is an integer and for any $\varepsilon < 1$, for arbitrarily large $x$, $f(x)$ is an integer minus $\varepsilon$. When $f(x)$ is an integer, $f(x) - \lfloor f(x) \rfloor = 0$ and when $f(x)$ is an integer minus $\varepsilon$, $f(x) - \lfloor f(x) \rfloor = \varepsilon$. So it is easy to see that $\lim_{x \rightarrow \infty} f(x) - \lfloor f(x) \rfloor$ doesn't exist.
Because $f(x) - \lfloor f(x) \rfloor$ is always non-negative and it is 0 for arbitrarily large $x$, its $\liminf$ is 0. Also, because $f(x) - \lfloor f(x) \rfloor$ is always less than 1 and is $\varepsilon$ for any $\varepsilon < 1$ for arbitrarily large $x$, its $\limsup$ is 1, as Yuval Filmus mentioned.
