calculus - How do I find this limit: $lim_{x to infty} sqrt{x^4-3x^2-1}-x^2$

$$

\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2
$$

The answer is
$$
\frac{-3}{2}
$$
according to Wolfram alpha.

Answer

I would consider $$\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$$

as $$\lim_{x \to \infty} \dfrac{\sqrt{x^4-3x^2-1}-x^2}{1}$$

Then multiply $\sqrt{x^4-3x^2-1}+x^2$ top and bottom to get rid of the ridiculous looking square root sign on the top. Then bingo!