Limit as x approaches negative infinity.

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Discussion Overview

The discussion revolves around the limit of a function as x approaches negative infinity, specifically examining the expression $$\lim_{{x}\to{-\infty}}\frac {x} {\sqrt{x^2}}$$ and the reasoning behind its evaluation. Participants explore the implications of the limit in relation to the sign function and the properties of square roots.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions the correctness of the limit evaluation, suggesting that if x is -1, the limit should yield 1 instead of -1.
  • Another participant clarifies that for x = -1, the expression evaluates to -1, and introduces the sign function as a relevant concept.
  • There is a reiteration of the sign function definition, indicating that it is defined as $$\text{sgn}(x)\equiv\frac{x}{|x|}$$ for $$x\ne0$$.
  • Participants discuss the relationship between the square root of squared values, with one stating that $$\sqrt{(-x)^2} = x$$, which is challenged by another who asserts that it equals $$|x|$$.
  • The absolute value definition is provided, showing that $$|x|$$ depends on whether x is negative or non-negative.

Areas of Agreement / Disagreement

Participants express differing views on the evaluation of the limit and the properties of square roots, indicating that there is no consensus on the interpretation of the limit or the definitions involved.

Contextual Notes

Participants have not resolved the assumptions regarding the behavior of the square root function and its relation to negative values, leading to potential misunderstandings in the limit evaluation.

tmt1
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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?
 
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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:

$$\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}$$
 
MarkFL said:
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 $$\sqrt{ (-x)^2} = x$$?
 
tmt said:
so $$\sqrt{ (-x)^2} = x$$?

No, what we have is:

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

Recall that:

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

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