Simplify $\frac{\sqrt{x^2 + x}}{x}$ How does this simplification work?
$\frac{\sqrt{x^2 + x}}{x} =\sqrt{1 + 1/x} $

$$\frac{\sqrt{x^2+x}}{x}=\frac{\sqrt{x^2+x}}{\sqrt{x^2}}=\sqrt{\frac{x^2+x}{x^2}}=\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}=\sqrt{1+\frac{1}{x}}$$
Note: this is valid for $x>0$, as then $x=|x|=\sqrt{x^2}$.
For $x<-1$, the LHS is defined, but negative, while the RHS is positive, so they do not agree.  For $-1<x<0$, the LHS is undefined as you're taking the square root of a negative number.  For $x=-1$, the two sides agree since they're both zero.  For $x=0$, neither side is defined since you're dividing by zero.
