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: The expression can be multiplied with its conjugate and then:
$$\begin{align}
\lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right) 
  &= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2-1}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} x\left(\frac{-2}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} \frac{-2}{\frac{\sqrt{x^2-1}}{x} + \frac{\sqrt{x^2+1}}{x}} \cr
  &=\lim_{x\to\infty} \frac{-2}{\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}} \cr
  &=\lim_{x\to\infty} \frac{-2}{\sqrt{1-0} + \sqrt{1-0}} \cr
  &=\lim_{x\to\infty} \frac{-2}{1+1} \cr
  &= -1\end{align}$$
A: Putting $\frac1{x^2}=h$
So, $h\to0$ as $x\to\infty$
$$\lim_{x\to\infty}x(\sqrt{x^2-1}-\sqrt{x^2+1})$$
$$=\lim_{h\to0}\frac{\sqrt{1-h}-\sqrt{1+h}}{h}$$
$$\text{Now, }\sqrt{1-h^2}-\sqrt{1+h^2}=\frac{(1-h^2)-(1+h^2)}{\sqrt{1-h^2}+\sqrt{1+h^2}}=\frac{-2h^2}{\sqrt{1-h^2}+\sqrt{1+h^2}}$$
$$\implies \frac{\sqrt{1-h}-\sqrt{1+h}}h=\frac{-2}{\sqrt{1-h}+\sqrt{1+h}}$$
$$\lim_{h\to0}\frac{\sqrt{1-h}-\sqrt{1+h}}h=\frac{-2}2=-1$$
Alternatively,
$$\lim_{h\to0}\frac{(1-h)^\frac12-(1+h)^\frac12}{h^2}$$
$$=-\lim_{h\to0}\frac{\left(1+\frac h2+O(h^2)\right)-\left(1-\frac h2+O(h^2)\right)}{h^2}$$
$$=-\lim_{h\to0}\frac{h+O(h^2)}h=-1$$
A: 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 
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)\sim_{+\infty}\frac{-2x}{2x}=-1.$$
