<s>[INST] The limit of a fraction What is the value of
$$\lim_{x\to\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}$$
I tried to factorise the denominator or the numerator but it takes me nowhere. My teacher said it equals to $\frac{1}{2}$. But I don't get it. Can anyone help me?
 [/INST] $$\lim_{x\rightarrow+\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow+\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$
Also
$$\lim_{x\rightarrow-\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow-\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$
Thus, indeed, the answer is $\frac{1}{2}$.
</s>