The conjugate trick at infinity

A tempting mistake. "For large $x$, $\sqrt{x^2+x}\approx x$, so the limit is $0$." This is wrong: the error in the approximation $\sqrt{x^2+x}\approx x$ is not small compared with the quantity we are measuring.

Computation. For $x>0$,
$$\sqrt{x^2+x}-x=\frac{(x^2+x)-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}+x}.$$
Divide the numerator and the denominator by $x$ (note $\sqrt{x^2+x}=x\sqrt{1+\frac1x}$ for $x>0$):
$$\frac{x}{\sqrt{x^2+x}+x}=\frac1{\sqrt{1+\frac1x}+1}\ \longrightarrow\ \frac1{1+1}=\frac12.$$

Remark. Geometrically, $\sqrt{x^2+x}$ is the distance $x+\tfrac12-\varepsilon$ with a tiny $\varepsilon$: the curve $y=\sqrt{x^2+x}$ has the line $y=x+\tfrac12$ as an asymptote. This is why a gap of $\tfrac12$ remains.