Question

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

Answer 1

Method of rationalising both numerators and denominators,

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

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

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

$=\underset{x\to 1}{lim}\frac{(\sqrt{x+8}-\sqrt{8x+1})(\sqrt{5-x}+\sqrt{7x-3})(\sqrt{x+8}+\sqrt{8x+1})}{-8(x-1)\left((\sqrt{x+8}+\sqrt{8x+1})\right)}$

$=\underset{x\to 1}{lim}\frac{(x+8-8x-1)(\sqrt{5-x}+\sqrt{7x-3})}{-8(x-1)\left((\sqrt{x+8}+\sqrt{8x+1})\right)}$

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

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

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

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

$=\frac{7(\sqrt{4}+\sqrt{4})}{8(\sqrt{9}+\sqrt{9})}$

$=\frac{14\sqrt{4}}{16\sqrt{9}}$

$=\frac{14\times 2}{16\times 3}$

$=\frac{7}{4\times 3}$

$=\frac{7}{12}$