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.

To write the complete solution, as per the hint in the comments I gave:
$$
\begin{aligned}
\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right) &= \lim_{x \to \infty}x^2\left(\sqrt{(x^2+3)^2} - \sqrt{x^4 + 6x^2}\right)\\
&=\lim_{x \to \infty}x^2\left(\frac{(x^2+3)^2-x^4-6x^2}{\sqrt{(x^2+3)^2} + \sqrt{x^4 + 6x^2}}\right)\\
&=\lim_{x \to \infty}\frac{9x^2}{x^2+3 + \sqrt{x^4 + 6x^2}}\\
&=\lim_{x \to \infty}\frac{9}{1+\frac{3}{x^2} + \sqrt{1 + \frac{6}{x^2}}}\\
&=\frac{9}{2}\\
\end{aligned}
$$
