<s>[INST] 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.
 [/INST] first, substitute : $t=x^2$, you get :
$$\lim_{t\to +\infty} \sqrt{t^2-3t-1}-t=\lim_{t\to \infty} \frac{-3t-1}{\sqrt{t^2-3t-1}+t}=\lim_{t\to \infty} \frac{-3-\frac{1}{t}}{\sqrt{1-\frac{3}{t} -\frac{1}{t^3}}+1} =\frac{-3}{2}. $$
Explanation : first we use this identity for $a\neq -b$ : $a-b =\frac{a^2-b^2}{a+b}$, then we factor $t$ from the numerator and the denomenator, the limit of the numerator is $-3$ and the limit of the numerator is $2$.

Further more, for any reals $a,b$ : $$\lim_{x\to+ \infty } \sqrt{x^2+ax+b}-x=\frac{a}{2}.$$
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