calculus - Solution of a limit of a sequence $frac{sqrt{4n^2+1}-2n}{sqrt{n^2-1}-n}$

I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:

$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$

Here there are my differents methods:

1. assuming $f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$, $h(n)=\sqrt{n^2-1}$, $\ \psi(n)= n$ $$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$
   I always have an undetermined form.


2. I've done some rationalizations:
   $$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$ where to the numerator I find $1$ and to the denominator an undetermined form. Similar situation considering
   $$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$


3. $$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$
   At the moment I am not able to think about other possible simple solutions.

Answer

Note
\begin{eqnarray}
\lim_{n\to\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}&=&\lim_{n\to\infty}\frac{(\sqrt{4n^2+1}-2n)(\sqrt{4n^2+1}+2n)}{(\sqrt{n^2-1}-n)(\sqrt{n^2-1}+n)}\cdot\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{1-\frac1{n^2}}+1}{\sqrt{4+\frac1{n^2}}+2}\\
&=&-\frac12.
\end{eqnarray}