How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seems to make it worse. Any help would be appreciated.

You have another solution using Taylor series. Rewrite $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=x^2\Big(\sqrt{1+a/x^2}-\sqrt{1+b/x^2}\Big)$$ and remember that, when $y$ is small $\sqrt{1+y}\simeq 1+y/2$. Replace successively $y$ by $a/x^2$ and $b/x^2$ and you are done.
If you continue in the same spirit you coulod show a beautiful thing $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=\frac{a-b}{2}-\frac{a^2-b^2}{8 x^2}+\frac{a^3-b^3}{16 x^4}-\frac{5
   \left(a^4-b^4\right)}{128 x^6}+O\left(\left(\frac{1}{x}\right)^7\right)$$
