Limits practice question 22

This limit runs to $-\infty$, so every $x$ in sight is negative. At which step of the usual divide-by-$x$ method does that sign matter?

Divide numerator and denominator by $x$. Inside the square root that means dividing by $x^{2}$, and here is the trap: for $x < 0$, $\sqrt{x^{2}} = -x$, not $x$.

So $\dfrac{\sqrt{9x^{2} + 1}}{x} = -\sqrt{9 + \tfrac{1}{x^{2}}}$ when $x$ is negative.