Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.

Just to give yet another approach, let $u=x^2+3$. Then, for $x\ge0$,
$$x^2(x^2-x\sqrt{x^2+6}+3)=(u-3)\left(u-\sqrt{u^2-9}\right)={9u\over u+\sqrt{u^2-9}}-{27\over u+\sqrt{u^2-9}}\to{9\over1+1}-0={9\over2}$$
