Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$

Question

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

Answer 1

I would say

$\displaystyle \lim_{x\to \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor=\lim_{x\to \infty} \lfloor \sqrt{x^2+x+1 }\rfloor+\{ \sqrt{x^2+x+1 } \}-\lfloor \sqrt{x^2+x+1 }\rfloor=$

$\displaystyle=\lim_{x\to \infty} \{ \sqrt{x^2+x+1 } \}= (0 \sim 1) \Rightarrow\, $limit not exist

Answer 2

As you mention, when $x$ is a large integer, the difference is roughly $1/2$. On the other hand, when $\sqrt{x^2+x+1}$ is an integer, which happens for unboundedly large $x$, the difference is $0$. This gives two different sequences tending to infinity with two different limits of your function $\sqrt{x^2+x+1}-\lfloor\sqrt{x^2+x+1}\rfloor$, showing that it doesn't tend to a limit. In fact, this shows that its $\liminf$ is $0$, and you can similarly show that the $\limsup$ is $1$.

Answer 3

$\forall a\in(0,1),n\in N^+$,let$x_n$ be the positive zero of : $$x^2+x+1=(n+a)^2$$ then $x_{n+1}>x_n$,and $x_n>n+a-1$,so$\lim_{n\to+\infty}x_n=+\infty$,but: $$\sqrt{x_n^2+x_n+1}-[x_n^2+x_n+1]\equiv a$$

Answer 4

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.