How can one find $\lim_{x \to \pm\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$ without graphing? I'm trying to find both of these:
$$\lim_{x \to +\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
$$\lim_{x \to -\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$

I know that they end up being $2$ and $-2$ by graphing, but what process can I use to find them without graphing (in the simplest way possible)?

First recall that $\sqrt{x^2} = x$ if $x\ge0$ and $\sqrt{x^2} = -x$ (which is positive) if $x<0.$
Then treat the two cases, $x>0$ and $x<0,$ separately:
\begin{align}
\frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt]
& = \frac{2+\frac 1 x}{\frac 1 {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x>0 \\[10pt]
& = \frac{2+ \frac 1 x}{\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt]
& \to 2 \text{ as } x\to\infty. \\[10pt]
\frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt]
& = \frac{2+\frac 1 x}{\frac {-1} {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x<0 \\[10pt]
& = \frac{2+ \frac 1 x}{-\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt]
& \to -2 \text{ as } x\to\infty.
\end{align}
