Question

Evaluate the limit:

$\underset{x\to 1}{lim}\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}$

Answer 1

At $x\to 1,\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}\text{becomes}\frac{0}{0}form$

$\underset{x\to 1}{lim}\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}$

On rationalising numerator, we get

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

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

$[\because (a-b\left)\right(a+b)=a^2-b^2]$

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

$=\underset{x\to 1}{lim}\frac{(4x-4)}{(x+1)(x-1)(\sqrt{5x-4}+\sqrt{x})}$

$=\underset{x\to 1}{lim}\frac{4(x-1)}{(x+1)(x-1)(\sqrt{5x-4}+\sqrt{x})}$

$\left[\right(x-1)\ne 0]$

$=\underset{x\to 1}{lim}\frac{4}{(x+1)(\sqrt{5x-4}+\sqrt{x})}$

$=\frac{4}{(1+1)(\sqrt{5\left(1\right)-4}+\sqrt{1})}$

$=\frac{4}{4}$

$=1$

$\therefore \underset{x\to 1}{lim}\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}=1$