<s>[INST] 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.
 [/INST] HINT
Multiply by a special kind of $1$  :
$$\lim\limits_{x \to \infty}\left[x(\sqrt {x^2+a} - \sqrt {x^2+b})\right]=\lim\limits_{x \to \infty}\left[ \dfrac{x(\sqrt {x^2+a} - \sqrt {x^2+b})(\sqrt {x^2+a} + \sqrt {x^2+b})}{\sqrt {x^2+a} + \sqrt {x^2+b}}\right] $$
and use difference of squares
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