TITLE: Find the limit $\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)$
QUESTION [9 upvotes]: $$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)\;=\;?\quad(n\in I) \\ \text{where $\lfloor\cdot\rfloor$ is the greatest integer function.}$$

This is what I did:
Since $[x] = x - \{x\}$ we get our limit equal to
$$\lim_{n\to\infty}\left\{\sqrt{n^2+n+1}\right\}$$
Moving the limit inside the fractional part function and replacing $n=\frac 1h \; \text {where } h\to0^+$ we get
$$\left\{\lim_{h\to0^+} \frac{\sqrt{h^2+h+1}}h\right\}$$
Applying L'Hospital Rule, we get our limit equal to $\left\{\frac 12\right\}$ which is $0$.

The problem:
The answer in the answer key is $\frac12$. So here, the only problem I seem to find in my solution is that $n\in I$ and simply assuming $n = \frac 1h$ doesn't ensure our $n$ to be an integer.
Can anyone provide a way to either correctly assume a new value for $n$ or any alternate way to solve this?

REPLY [6 votes]: Alternative idea for proving the essential facts: writing
$$\sqrt{n^2+n+1} = n\sqrt{1+1/n+1/n^2}$$
and using the Taylor series of $\sqrt{1+x}$:
$$1+{\frac{x}2}-{\frac{x^2}8}+O(x^3)$$
we have
$$\sqrt{n^2+n+1} = n+\frac12+\frac3{8n}+O(1/n^2)$$
and
$$\lfloor\sqrt{n^2+n+1}\rfloor = n.$$