Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$

The expression can be multiplied with its conjugate and then:

$$\begin{align} \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right) &= \lim_{x\to\infty} x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)\left(\frac{\sqrt{x^2-1}+\sqrt{x^2+1}}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{x^2-1-x^2-1}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} x\left(\frac{-2}{\sqrt{x^2-1}+\sqrt{x^2+1}}\right) \cr &=\lim_{x\to\infty} \frac{-2}{\frac{\sqrt{x^2-1}}{x} + \frac{\sqrt{x^2+1}}{x}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}} \cr &=\lim_{x\to\infty} \frac{-2}{\sqrt{1-0} + \sqrt{1-0}} \cr &=\lim_{x\to\infty} \frac{-2}{1+1} \cr &= -1\end{align}$$

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,

$$\lim_{h\to0}\frac{(1-h)^\frac12-(1+h)^\frac12}{h^2}$$

$$=-\lim_{h\to0}\frac{\left(1+\frac h2+O(h^2)\right)-\left(1-\frac h2+O(h^2)\right)}{h^2}$$

$$=-\lim_{h\to0}\frac{h+O(h^2)}h=-1$$