limit of: $\begin{align} \lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n} \end{align}$ How to get the limit of:
$\begin{align}
\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}
\end{align}$
I have formed it to: 
$\begin{align}
\lim_{n\to \infty}\frac{n(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{n}
\end{align}$
And tried to use L'Hopital but unfortunately I am not able to get a result with that:
$f'(x)=\frac{d}{dn}(n(\sqrt{n-2\sqrt{n}}-\sqrt{n}))= (\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left(  \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}}  - \frac{1}{2\sqrt{n}}\right)$
$g'(x)=\frac{d}{dn} n = 1$
$\begin{align}
\lim_{n \to \infty}\frac{f'(x)}{g'(x)} &=
\lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left(  \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}}  - \frac{1}{2\sqrt{n}}\right)}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( 0\right)}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{1}} \\
&= \lim_{n \to \infty}{\frac{(\sqrt{\frac{1}{n}-2\sqrt{\frac{1}{n}}}-\sqrt{\frac{1}{n}})}{\frac{1}{n}}} \\
&= 0\\
\end{align}$
Which is wrong. Wolfram says that the limit is $-1$
I appreciate every help :)

$$\lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n-2\sqrt{n}}+\sqrt{n}}=\lim_{n\to \infty}\frac{-2\sqrt{n}}{\sqrt{n}-1+\sqrt{n}}=-1$$
