Question

solve the trigonometric equation
$l{t}_{x\to 1}\frac{\sqrt{x+8}-\sqrt{7x+2}}{\sqrt{x+3}-\sqrt{5-x}}$

Answer 1

Rationalising the numerator and denominator, we get

${lim}_{x\to 1}\frac{\sqrt{x+8}-\sqrt{7x+2}}{\sqrt{x+3}-\sqrt{5-x}}\times \frac{\sqrt{x+3}+\sqrt{5-x}}{\sqrt{x+3}+\sqrt{5-x}}\times \frac{\sqrt{x+8}+\sqrt{7x+2}}{\sqrt{x+8}+\sqrt{7x+2}}$

$={lim}_{x\to 1}\frac{(\sqrt{x+8}-\sqrt{7x+2})(\sqrt{x+3}+\sqrt{5-x})(\sqrt{x+8}+\sqrt{7x+2})}{(\sqrt{x+3}-\sqrt{5-x})(\sqrt{x+3}+\sqrt{5-x})(\sqrt{x+8}+\sqrt{7x+2})}$

$={lim}_{x\to 1}\frac{(x+8-7x-2)(\sqrt{x+3}+\sqrt{5-x})}{(x+3-5+x)(\sqrt{x+8}+\sqrt{7x+2})}$

$={lim}_{x\to 1}\frac{(6-6x)(\sqrt{x+3}+\sqrt{5-x})}{(2x-2)(\sqrt{x+8}+\sqrt{7x+2})}$

$={lim}_{x\to 1}\frac{6(1-x)(\sqrt{x+3}+\sqrt{5-x})}{2(x-1)(\sqrt{x+8}+\sqrt{7x+2})}$

$={lim}_{x\to 1}\frac{-3(\sqrt{x+3}+\sqrt{5-x})}{(\sqrt{x+8}+\sqrt{7x+2})}$

$=\frac{-3(\sqrt{1+3}+\sqrt{5-1})}{(\sqrt{1+8}+\sqrt{7+2})}$

$=\frac{-3(\sqrt{4}+\sqrt{4})}{(\sqrt{9}+\sqrt{9})}$

$=\frac{-3(2+2)}{(3+3)}$

$=\frac{-3\times 4}{2\times 3}=-2$