What's wrong about this limit?

Question

i've got a very simple exercise of limit, and i started solving it, but it won't get out of indetermination.

The limit is $$\lim_{x\to 1} = \frac{x^2-\sqrt x}{1-\sqrt x}$$

If i use L'Hôpital's it works, but just using basic math rules, it falls in another indetermination ($\frac{0}{0}$).

Answer 1

$$\frac{x^2-\sqrt x}{1-\sqrt x}=\frac{(x^2-\sqrt x)(1+\sqrt x)}{1-x}\;\color{red}{(**)}$$

But also

$$x^2-\sqrt x=\frac{x^4-x}{x^2+\sqrt x}=x\frac{\overbrace{(x-1)(x^2+x+1)}^{=x^3-1}}{x^2+\sqrt x}$$

so

$$\color{red}{(**)}=-\frac{x(x^2+x+1)(1+\sqrt x)}{x^2+x}\xrightarrow[x\to1]{}-\frac{3\cdot2}{2}=-3$$

Directly with l'Hospital:

$$\lim_{x\to1}\frac{x^2-\sqrt x}{1-\sqrt x}=\lim_{x\to1}\frac{2x-\frac1{2\sqrt x}}{-\frac1{2\sqrt x}}=\lim_{x\to1}\left(-4x^{3/2}+1\right)=-3$$

Answer 2

Notice that:

$$\frac{x^2-\sqrt{x}}{1-\sqrt{x}}=\frac{\left(x^2-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=$$ $$-\sqrt{x}\left(x\sqrt{x}+1\right)=-\sqrt{x}-x-x^{\frac{3}{2}}$$

Answer 3

if $t=\sqrt x$

$\lim_{x\to1}\frac{x^2-\sqrt x}{1-\sqrt x}=\lim_{t\to1}\frac{t^4-t}{1-t}=\lim_{t\to1}\frac{(t^3+t^2+t)(t-1)}{1-t}=-3$