Question

Calcule o $\lim_{x \to \ 4} \frac{ 3-\sqrt{5+x} }{ 1-\sqrt{5-x} }$

Answer 1

$\lim_{x\to4}\frac{(3-\sqrt{5+x})}{(1-\sqrt{5-x})}=\\\\\\\lim_{x\to4}\frac{(3-\sqrt{5+x})}{(1-\sqrt{5-x})}\times\frac{(3+\sqrt{5+x})}{(3+\sqrt{5+x})}\times\frac{(1+\sqrt{5-x})}{(1+\sqrt{5-x})}=\\\\\\\lim_{x\to4}\frac{(3-\sqrt{5+x})}{(1-\sqrt{5-x})}\times\frac{(3+\sqrt{5+x})}{(1+\sqrt{5-x})}\times\frac{(1+\sqrt{5-x})}{(3+\sqrt{5+x})}=\\\\\\\lim_{x\to4}\frac{(9-5-x)}{(1-5+x)}\times\frac{(1+\sqrt{5-x})}{(3+\sqrt{5+x})}=\\\\\\\lim_{x\to4}\frac{(4-x)}{(-4+x)}\times\frac{(1+\sqrt{5-x})}{(3+\sqrt{5+x})}=$

$\lim_{x\to4}\frac{(4-x)}{-1\cdot(4-x)}\times\frac{(1+\sqrt{5-x})}{(3+\sqrt{5+x})}=\\\\\\\lim_{x\to4}\frac{(1+\sqrt{5-x})}{-(3+\sqrt{5+x})}=\\\\\\\frac{1+\sqrt{5-4}}{(3+\sqrt{5+4}})=\\\\\\\frac{1+1}{3+3}=\\\\\\\boxed{\frac{1}{3}}$