Where did $\sqrt{x^2/x^2}$ come from in $\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} = \lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1$?

Question

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.

Answer 1

I guess it's an exercise meant to show how things might go wrong without being careful doing algebraic manipulations.

Since they're evaluating the limit at $-\infty$, it's not restrictive to assume $x<0$. Divide numerator and denominator by $-x$: $$ \frac{x+1}{\sqrt{x^2}}= \frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}} $$ Since $x<0$, we have $-x>0$, so $-x=\sqrt{(-x)^2}=\sqrt{x^2}$ and so $$ \frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}} = \frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{\sqrt{x^2}}} = \frac{-1-\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}}}= -1-\frac{1}{x} $$ A very common error would be dividing numerator and denominator by $x$ and bringing $x$ in the square root: $$ \frac{x+1}{\sqrt{x^2}}= \frac{1+\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{x}} \color{red}{=} \frac{1+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}}} $$ where the red equals sign is wrong.

On the other hand, recalling that $\sqrt{x^2}=|x|=-x$ (because we're in the interval $x<0$), the transformation is much simpler: $$ \frac{x+1}{\sqrt{x^2}}= \frac{x+1}{-x}=-1-\frac{1}{x} $$

Answer 2

The reasoning seems to be, for $x<0$, $$ \sqrt{x^2}\cdot(-1/x)=\sqrt{x^2}\cdot(1/|x|)=\sqrt{x^2}\cdot(1/\sqrt{x^2})=\sqrt{x^2/x^2}. $$

Answer 3

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 $$\sqrt{x^2}\times\left(-\frac 1x\right)=\sqrt{x^2}\times\frac{1}{-x}=\sqrt{x^2}\times \frac{1}{\sqrt{(-x)^2}}=\sqrt{\frac{x^2}{x^2}}$$