How to calculate $\lim_{x\to 1} \frac{\sqrt x - x^2}{1-\sqrt x}$ without L'Hopital rule?

Question

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

I solved it by using L'hopital and got $3$.
I've also attempted to solve it without L'hopital, but unfortunately I couldn't.

Answer 1

Hint: Use $$\sqrt{x}-x^2=\sqrt{x}(1-\sqrt{x}^3)=\sqrt{x}(1-\sqrt{x})(1+\sqrt{x}+x)$$

Answer 2

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

Answer 3

Hint: After taking $\sqrt{x}$ common, use the formula $(a+b)^3=(a+b)(a^2-ab+b^2)$ for $a=1,b=-x^{1/2}$.