Where did $\sqrt{x^2/x^2}$ come from in $\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} = \lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1$? I'm reading a calculus book and I saw the following limit solution.
$$
\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} =
\lim_{x \to -\infty}
  \left(\frac{x+1}{\sqrt{x^2}} \cdot \frac{-1/x}{-1/x}\right) =
\lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1.
$$
I'm having some trouble to understand where did the $\sqrt{x^2/x^2}$ came from.

The solution in the book multiplies $(x+1)/\sqrt{x^2}$ by $(-1/x)/(-1/x)\ (=1)$.
Since $x\lt 0$, the denominator will be
$$\sqrt{x^2}\times\left(-\frac 1x\right)=\sqrt{x^2}\times\frac{1}{-x}=\sqrt{x^2}\times \frac{1}{\sqrt{(-x)^2}}=\sqrt{\frac{x^2}{x^2}}$$
