Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$
$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{3+\text{x}}-1}{2-\text{x}}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{3-\text{x}}-1\big)}{2-\text{x}}\times\frac{\big(\sqrt{3-\text{x}}+1\big)}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(3-\text{x})-1}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(2-\text{x})}{(2-\text{x})\big(\sqrt{3-\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\big(\sqrt{3-\text{x}}+1\big)}$
$=\frac{1}{\sqrt{3-2}+1}=\frac{1}{1+1}$
$=\frac{1}{2}$
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$\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$