Question

$\underset{x\to 1}{\lim }\frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1}$

Answer 1

$\underset{x\to 1}{\lim }\left[\frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator:

$\underset{x\to 1}{\lim }\left[\frac{\left(\sqrt{5x-4}-\sqrt{x}\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}{\left(x^3-1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

= $\underset{x\to 1}{\lim }\left[\frac{5x-4-x}{\left(x-1\right)\left(x^2+x+1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

= $\underset{x\to 1}{\lim }\left[\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

= $\frac{4}{3\left(1+1\right)}$

= $\frac{2}{3}$