<s>[INST] Another limit task, I've multiplied by the conjugate, now what? The task: 
$\lim_{x\to\infty} \sqrt{x^2+1} -x $
I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this
$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $
Is this correct so far? And what would be the next step? 
 [/INST] We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$
</s>