<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] A common trick: multiply by $1$.
That is: multiply $x(\sqrt{x^2+a^2}-\sqrt{x^2+b^2})$ by $\frac{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$
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