# Evaluating $\lim_{n \to \infty}\left(\sqrt{n^2 - n+1}-\left\lfloor\sqrt {n^2 - n+1}\right\rfloor\right)$

How do I evaluate $$\lim_{n\to\infty}\left(\sqrt{n^2-n+1}-\left\lfloor\sqrt{n^2 - n+1}\right\rfloor\right),n\in\Bbb N$$

Attempt:

I thought of using Squeeze theorem but that could not help.

Secondly, we know that $$x- \lfloor x\rfloor=\{x\}$$ where $$\{\}$$ denotes the fractional part function. But I am not sure how to actually evaluate limits involving the fractional part function.

• Hint: $n - 1 < \sqrt{n^2 - n + 1} < n$. Oct 2, 2018 at 13:34
• "But I am not sure how to actually evaluate limits involving the fractional part function." One common way is to switch out $\{x\}$ with $x-[x]$, and then... Oh, wait. Oct 2, 2018 at 13:39

Since $$n - 1 < \sqrt{n^2 - n + 1} < n$$, then\begin{align*} &\mathrel{\phantom{=}}{} \sqrt{n^2 - n + 1} - [\sqrt{n^2 - n + 1}] = \sqrt{n^2 - n + 1} - (n - 1)\\ &= \frac{n}{\sqrt{n^2 - n + 1} + (n - 1)} → \frac{1}{2}. \quad (n → ∞) \end{align*}
• As $n\to\infty$, we can assume that $n>1$, then $(n-1)^2=n^2-2n+1<n^2-n+1<n^2$. Take the square root and we arrive at the first inequality. If we take $n\in\mathbb Z$, then we can use the definition of the floor function to write $[\sqrt{n^2-n+1}]=n-1$. Oct 2, 2018 at 14:16