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.

x>0;
$f(x):=\sqrt{x^2+6}= x\sqrt{1+6/x^2}=$
$x(1+3/x^2+$
$(1/2)(-1/2)(1/2!)(6/x^2)^2+ O(1/x^6))=$
$x(1+3/x^2-(6^2/8)/x^4+O(1/x^6));$
$x^2(x^2-xf(x)+3)=$
$x^2(- 3+3+(9/2)/x^2+O(1/x^4))$;
Take the limit.
