Question

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

Answer 1

$\underset{x\to 1}{\lim }\left[\frac{x^2-\sqrt{x}}{\sqrt{x}-1}\right]$

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

$\underset{x\to 1}{\lim }\left[\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{\sqrt{x}-1}\right]$

= $\underset{x\to 1}{\lim }\left[\frac{\sqrt{x}\left(x^{3/2}-1\right)}{x^{{\displaystyle \frac{1}{2}}}-1}\right]$

= $\underset{x\to 1}{\lim }\left[\frac{\sqrt{x}\left({\left(\sqrt{x}\right)}^3-1^3\right)}{\left(\sqrt{x}-1\right)}\right]\left[A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\right]$

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

= 1 (1 + 1 + 1)

= 3