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When proving that $\lim_{x\to 2}(\sqrt{x+2}-\sqrt{x+1})=2-\sqrt3$ with $ε-δ$, if we multiply and divide by $\sqrt{x+2}+\sqrt{x+1}-(2-\sqrt3)$, we get:

$|f(x)-l|=\frac{|\sqrt{x+2}-\sqrt{x+1}-(2-\sqrt3)||\sqrt{x+2}+\sqrt{x+1}-(2-\sqrt3)|}{|\sqrt{x+2}+\sqrt{x+1}-(2-\sqrt3)|}=....=\frac{|4(2-\sqrt3)^2(x+2)-48|}{|\sqrt{x+2}+\sqrt{x+1}-(2-\sqrt3)||2(2-\sqrt3)\sqrt{x+2}-4\sqrt3|}$

In which I don't see the term $|x-2|$ appearing anywhere so it seems I must be following the wrong path? Is there a simpler solution to this?

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    Perhaps working with $x-2=t$ is simpler than this, at last set $t=x-2$. – 2017-02-16

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Observe that $$\sqrt{x+2}-2=\frac{x-2}{\sqrt{x+2}+2}$$ and $$\sqrt{x+1}-\sqrt{3}=\frac{x-2}{\sqrt{x+1}+\sqrt{3}}$$

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    Yes. I like. $$|\sqrt{x+2}-2-(\sqrt{x+1}-\sqrt{3})|\leq\frac{|x-2|}{\sqrt{x+2}+2}+\frac{|x-2|}{\sqrt{x+1}+\sqrt{3}}\leq2\epsilon$$ – 2017-02-16