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.

The expression can be multiplied with its conjugate and then:
$$\begin{align}
\lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right) 
  &= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2-1}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} x\left(\frac{-2}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr
  &=\lim_{x\to\infty} \frac{-2}{\frac{\sqrt{x^2-1}}{x} + \frac{\sqrt{x^2+1}}{x}} \cr
  &=\lim_{x\to\infty} \frac{-2}{\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}} \cr
  &=\lim_{x\to\infty} \frac{-2}{\sqrt{1-0} + \sqrt{1-0}} \cr
  &=\lim_{x\to\infty} \frac{-2}{1+1} \cr
  &= -1\end{align}$$
