Question

Evaluate
$\rm \displaystyle \lim_{x\to 1}\ \frac{\sqrt{5x-4}-\sqrt{5-4x}}{x^{2}-1}$

Answer 1

To solve this limit,first rationalising the numerator

$\lim_{x\to 1}\ \frac{\sqrt{5x-4}-\sqrt{5-4x}}{x^{2}-1} \\ \\ = \lim_{x\to 1}\ \frac{\sqrt{5x-4}-\sqrt{5-4x}}{x^{2}-1} \times \frac{ \sqrt{5x - 4} + \sqrt{5 - 4x} }{ \sqrt{5x - 4} + \sqrt{5 - 4x} } \\ \\ \lim_{x\to 1}\ \frac{5x-4-5-4x}{(x-1)(x + 1)(\sqrt{5x - 4} + \sqrt{5 - 4x})} \\ \\ = \lim_{x\to 1}\ \frac{9x-9}{(x-1)(x + 1)(\sqrt{5x - 4} + \sqrt{5 - 4x})} \\ \\ = \lim_{x\to 1}\ \frac{9(x-1)}{(x-1)(x + 1)(\sqrt{5x - 4} + \sqrt{5 - 4x})} \\ \\ =\lim_{x\to 1}\ \frac{9}{(x + 1)(\sqrt{5x - 4} + \sqrt{5 - 4x})} \\ \\ apply \: limit \\ \\ = \frac{9}{(1 + 1)(\sqrt{5(1) - 4} + \sqrt{5 - 4(1)})} \\ \\ = \frac{9}{2 \times 2} \\ \\ \lim_{x\to 1}\ \frac{\sqrt{5x-4}-\sqrt{5-4x}}{x^{2}-1} = \frac{9}{4}$
Hope it helps you.