How to find $ \lim _{x\to \infty }\left\{\sqrt{x^2+x+1}\right\} $ where $\{x\}=x-[x] =$ fractional part?

Question

$$ \lim _{x\to \infty }\left\{\sqrt{x^2+x+1}\right\} $$

where $\{x\}=x-[x] =$ fractional part

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I've got not determination case infinit - infinit and don't know what other route to take

Answer 1

Note that for any positive integer $x$, we have $$ x = \sqrt{x^2} \leq \sqrt{x^2 + x + 1} \leq \sqrt{x^2 + 2x + 1} = x+1 $$ Thus, we have $$ \left\{\sqrt{x^2 + x + 1}\right\} = \sqrt{x^2 + x + 1} - x $$ we then find $$ \sqrt{x^2 + x + 1} - x = \frac{x+1}{\sqrt{x^2 + x + 1} + x} \to \frac 12 $$ which is the desired limit.

Answer 2

I guess that here $x$ represents a real variable (if $x$ is an integer variable see Omnomnomnom's answer).

For any positive integer $n>1$, let $x_n$ be the positive real number such that $x_n^2+x_n+1=n^2$ (note that $f(x)=x^2+x+1$ is a continuous positive function such that $\lim_{x\to+\infty}f(x)=+\infty$).

Then along this sequence: $$\left\{\sqrt{x_n^2+x_n+1}\right\}=\left\{n\right\}=0.$$

Similarly, for any positive integer $n>1$, let $y_n$ be the positive real number such that $y_n^2+y_n+1=(n+1/2)^2$. Then along this sequence: $$\left\{\sqrt{y_n^2+y_n+1}\right\}=\left\{n+1/2\right\}=1/2.$$

What may we conclude?