Question

Evaluate the limit:
$\underset{x\to 1}{lim}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}$

Answer 1

At $x\to 1,$

$\frac{x^2-\sqrt{x}}{\sqrt{x}-1}$ becomes $\frac{0}{0}$ form

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

$=\underset{x\to 1}{lim}\frac{\sqrt{x}(x^{\frac{3}{2}}-1)}{\sqrt{x}-1}$

$=\underset{x\to 1}{lim}\frac{\sqrt{x}\left(\right(\sqrt{x}{)}^3-1)}{\sqrt{x}-1}$

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

$[\sqrt{x}-1\ne 0]$

$=\underset{x\to 1}{lim}\sqrt{x}(x+\sqrt{x}+1)$

$=\sqrt{1}(1+\sqrt{1}+1)=3$