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

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?
 A: The crux of this is that $\sqrt{n^2+n+1}\approx n+\frac{1}{2}$ for large $n$. 
Note that $$n^2+n+1=\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\implies\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}$$
One can thus quickly see that for large enough $n$, $n<\sqrt{n^2+n+1}<n+1$, so $\left\lfloor\sqrt{n^2+n+1}\right\rfloor=n$ eventually.
Thus we can see that \begin{align}\sqrt{n^2+n+1}-\lfloor \sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n\\ &=\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)+\frac{1}{2}\\ &=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}+\frac{1}{2}\\ &\to\frac{1}{2}\end{align}
A: 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
$$\lfloor\sqrt{n^2+n+1}\rfloor = n.$$
A: For every natural number, we have
$$\lfloor \sqrt{n^2+n+1} \rfloor =n$$
because $n^2\leq n^2+n+1< n^2+2n+1=(n+1)^2$. So we get
$$\begin{align}\lim_{n\to\infty} (\sqrt{n^2+n+1}- \lfloor \sqrt{n^2+n+1} \rfloor)&=\lim_{n\to\infty}(\sqrt{n^2+n+1}-n)\\
&=\lim_{n\to\infty}\frac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\\
&=\lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n+1}+n}\\
&=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\\
&=\frac{1}{2}
\end{align}$$
