MHB Limit as x approaches negative infinity.

[tmt1]

For this function:

$$\lim_{{x}\to{-\infty}}\frac {x} {\sqrt{x^2}} = -1$$

Why is this correct?

If x is equal to -1, for example, -1 square is 1. And the square root of 1 is 1. So should the answer be 1?

 

[MarkFL]

If $x=-1$, then you have:

$$\frac{-1}{1}=-1$$

The expression in the limit is one way to define the sign function:

$$\text{sgn}(x)\equiv\frac{x}{|x|}$$ where $$x\ne0$$

This is equivalent to the piecewise definition:

$$\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}$$

 

[tmt1]

If $x=-1$, then you have:

$$\frac{-1}{1}=-1$$

The expression in the limit is one way to define the sign function:

$$\text{sgn}(x)\equiv\frac{x}{|x|}$$ where $$x\ne0$$

This is equivalent to the piecewise definition:

$$\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}$$

so $$\sqrt{ (-x)^2} = x$$?

 

[MarkFL]

so $$\sqrt{ (-x)^2} = x$$?

No, what we have is:

$$\sqrt{(-x)^2}=\sqrt{x^2}\equiv|x|$$

Recall that:

$$|x|\equiv\begin{cases}-x, & x<0 \\[3pt] x, & 0\le x \\ \end{cases}$$