Horizontal asymptote of a radical rational function

Since $\sqrt{x^4} = x^2$ for all real $x$ (note $x^2 \ge 0$ always), we can divide numerator and denominator by $x^2$: $g(x) = \frac{x^2+1}{\sqrt{x^4+x+1}} = \frac{(x^2+1)/x^2}{\sqrt{x^4+x+1}/x^2} = \frac{1 + 1/x^2}{\sqrt{(x^4+x+1)/x^4}} = \frac{1+1/x^2}{\sqrt{1 + 1/x^3 + 1/x^4}}.$ This division by $x^2$ works identically whether $x \to +\infty$ or $x \to -\infty$, since $x^2$ is positive in both cases.

As $x \to \pm\infty$, the terms $1/x^2$, $1/x^3$, and $1/x^4$ all approach $0$. Substituting these limits gives $\lim_{x\to\pm\infty} g(x) = \frac{1+0}{\sqrt{1+0+0}} = \frac{1}{1} = 1.$ Since both one-sided limits at infinity equal $1$, the graph of $g$ has a single horizontal asymptote $y = 1$ (approached in both directions).