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I need to find the limit $$ \displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right),$$ where $n\in \mathbb{N}$.

My attempt. As $\displaystyle \lim_{n\rightarrow \infty} (n^2+n+1)\approx n^2$, then $\displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2+n+1}\approx \displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2} = n$.

So $$\displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-n\right) = \displaystyle \lim_{n\rightarrow \infty}\frac{\left(\sqrt{n^2+n+1}-n\right).\left(\sqrt{n^2+n+1}+n\right)}{\left(\sqrt{n^2+n+1}+n\right)}.$$

So $$\displaystyle \lim_{n\rightarrow \infty}\frac{n\cdot\left(1+\frac{1}{n}\right)}{n \left(\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1\right)} = \frac{1}{2}.$$

My Question is , Is my Process is Right OR Not ,OR Is there is any error .

If Not Then How can I Solve it

Help Required

Thanks

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    $\begingroup$ You wrote one limit in the title, yet did another one in the body of the question, so which one is it? $\endgroup$ – DonAntonio Oct 10 '13 at 8:41
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    $\begingroup$ $\lfloor \sqrt{n^2+n+1} \rfloor = n$, since $n^2 < n^2+n+1 < (n+1)^2$. $\endgroup$ – njguliyev Oct 10 '13 at 8:42
  • $\begingroup$ actually only one limit, Is was done all my mistake. Thanks njguliyev $\endgroup$ – juantheron Oct 10 '13 at 8:43
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Note that $$ n^2<n^2+n+1<n^2+2n+1=(n+1)^2, $$ and hence $$ n<\sqrt{n^2+n+1}<n+1, $$ and therefore $\lfloor\sqrt{n^2+n+1}\rfloor=n$. Therefore \begin{align} \sqrt{n^2+n+1}-\lfloor\sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n =\frac{\big(\sqrt{n^2+n+1}-n\big)\big(\sqrt{n^2+n+1}+n\big)}{\big(\sqrt{n^2+n+1}+n\big)}\\ &= \frac{n+1}{\sqrt{n^2+n+1}+n}=\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\to\frac{1}{2}, \end{align} as $n\to \infty$.

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