<s>[INST] Calculating limits with Strange Rearrangement How did they get to this third step?
$\lim\limits_{x \to 3} \frac{\sqrt{x+1}-2}{\sqrt{x-2}-1}$ =
$\lim\limits_{x \to 3} \frac{(\sqrt{x+1}-2)(\sqrt{x-2}+1)}{x-3}$ =
$\lim\limits_{x \to 3} \frac{(x-3)(\sqrt{x+1}-2)}{(x-3)(\sqrt{x+1}-2)}$
 [/INST] $\frac{\sqrt{x+1}-2}{\sqrt{x-2}-1} \cdot \frac{\sqrt{x+1}+2}{\sqrt{x-2}+1} \cdot \frac{\sqrt{x-2}+1}{\sqrt{x+1}+2} \\ \text{ multiply the first two fractions } \\ \text{ Also I multiplied the initial fraction you have by one } $
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