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.

$$x^2(x^2+6)=(x^2+3)^2-3^2$$
WLOG $x^2+3=3\csc t,t\to0^+$
$$\lim_{t\to0^+}3(\csc t-1)(3\csc t-3\cot t)$$
$$=\lim\dfrac{9(1-\sin t)(1-\cos t)}{\sin^2t}$$
$$=9\lim\dfrac{1-\sin t}{1+\cos t}=?$$ as $1-\cos t\ne0$ as $t\ne0$ as $\to0$
