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.

If you write the expression as $$x^4\left(1-\sqrt{1+6/x^2}+3/x^2\right)=\frac{1-\sqrt{1+6/x^2}+3/x^2}{1/x^4},$$ then you may be able to apply the Marquis de L'hopital's method.
After exactly two iterations, you get $$\frac92\frac{\frac{1}{\sqrt{1+6/x^2}}-1}{1/x^2},$$ where you may now take an elementary limit as $x\to+\infty.$
