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.
 A: Hint: Write the terms as a difference $a-b$, then multiply with the conjugate $a+b$ 
A: To write the complete solution, as per the hint in the comments I gave:
$$
\begin{aligned}
\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right) &= \lim_{x \to \infty}x^2\left(\sqrt{(x^2+3)^2} - \sqrt{x^4 + 6x^2}\right)\\
&=\lim_{x \to \infty}x^2\left(\frac{(x^2+3)^2-x^4-6x^2}{\sqrt{(x^2+3)^2} + \sqrt{x^4 + 6x^2}}\right)\\
&=\lim_{x \to \infty}\frac{9x^2}{x^2+3 + \sqrt{x^4 + 6x^2}}\\
&=\lim_{x \to \infty}\frac{9}{1+\frac{3}{x^2} + \sqrt{1 + \frac{6}{x^2}}}\\
&=\frac{9}{2}\\
\end{aligned}
$$
A: $$x^2(x^2+6)=(x^2+3)^2-3^2$$
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 $\to0$
A: Set $1/x=h$ to find
$$\lim_{h\to0^+}\dfrac{1-\sqrt{1+6h^2}+3h^2}{h^4}$$
$$=\lim\dfrac{(1+3h^2)^2-(1+6h^2)}{h^4}\cdot\lim\dfrac1{1+3h^2+\sqrt{1+6h^2}}$$ rationalizing the numerator
$$=\dfrac9{1+\sqrt1}$$
A: Just to give yet another approach, let $u=x^2+3$. Then, for $x\ge0$,
$$x^2(x^2-x\sqrt{x^2+6}+3)=(u-3)\left(u-\sqrt{u^2-9}\right)={9u\over u+\sqrt{u^2-9}}-{27\over u+\sqrt{u^2-9}}\to{9\over1+1}-0={9\over2}$$
A: 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.
A: 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 $x\to+\infty.$
