Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$ How can I show/explain the following limit?

$$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$

Some trivial transformation seems to be eluding me.

A short calculation, using equivalents:
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)=\frac{x\bigl((x^2-1)-(x^2+1)\bigr)}{\sqrt{x^2-12}+\sqrt{x^2+1}}= \frac{-2x}{\sqrt{x^2-1}+\sqrt{x^2+1}}.$$
Now, for $x>0$, we have
$$ \sqrt{x^2-1}+\sqrt{x^2+1}=x\biggl(\sqrt{1-\frac1{x^2}}+\sqrt{1+\frac 1{x^2}\biggr)}
\sim_{+\infty}2x, $$
since the contents of the parenthesis tends to $2$ at $\infty$, so 
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)\sim_{+\infty}\frac{-2x}{2x}=-1.$$
