Basic limit involving absolute value

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression \(\frac{|x^{2}-1|}{x^{2}+x}\) as \(x\) approaches -1. The problem involves concepts related to limits and absolute values in calculus.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating one-sided limits to handle the absolute value, with some attempting to express the limit as a piecewise function. There are questions about the signs of the limits from both sides and the implications of the absolute value.

Discussion Status

The conversation is ongoing, with participants sharing their calculations and questioning each other's reasoning. Some guidance has been offered regarding the treatment of absolute values and the necessity of considering one-sided limits, but no consensus has been reached on the correct approach or results.

Contextual Notes

Participants are grappling with the implications of the absolute value in the limit and the behavior of the function around the point of interest. There is mention of potential sign errors and the need for clarification on the conditions under which the absolute value changes.

wvcaudill2
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Homework Statement


lim as x[itex]\rightarrow[/itex]-1 = [itex]\frac{|x^{2}-1|}{x^{2}+x}[/itex]


Homework Equations


N/a


The Attempt at a Solution


Tried to write this as a piecewise function, but I got lost.
 
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Hi wvcaudill2! :smile:

Calculate the right-sided and the left-sided limits separately. That way, you could get rid of those nasty absolute value signs!
 
Ok, from the left-hand side of the limit, I got -2, and from the right-hand side I got 2, and when x-1=0 I got "does not exist." Is this right?
 
wvcaudill2 said:
Ok, from the left-hand side of the limit, I got -2, and from the right-hand side I got 2,

The correct solution should be those two switched. From the left side, you should get 2. From the right side, you should get -2. Check your signs.

and when x-1=0 I got "does not exist."

What in Earth do you mean with this??
 
I do not see how they should be switched. Can you explain your process in more detail?
 
wvcaudill2 said:
I do not see how they should be switched. Can you explain your process in more detail?

Well, show your work, and I'll show you what went wrong...
 
eqp1.jpg
 
You can't just eliminate the x+1 in your first line. Why not? Because it sits inside an absolute value sign. This will cause sign errors.

The absolute first thing you should do is eliminate the absolute value signs around [itex]x^2-1[/itex]. If x comes from the left, then this expression is positive, so you can drop the absolute valuie signs. If x comes from the right, then it is negative, so you'll have to write a minus sign in front.
 
Im still not sure how it is you are coming to the statement that "if x comes from the left, then this expression is positive." Can you explain how you are getting this?

Here is what I have now:
eqp2.jpg
 
  • #10
wvcaudill2 said:
Im still not sure how it is you are coming to the statement that "if x comes from the left, then this expression is positive." Can you explain how you are getting this?

Here is what I have now:
eqp2.jpg

Well, if [itex]x\leq -1[/itex], then [itex]x^2-1\geq 0[/itex]. This is easily seen by drawing the parabola for [itex]x^2-1[/itex].
 
  • #11
Ok, I see what you are saying now, but when I graphed the parabola, I also saw that coming from the right-hand side, the values of x were also positive up until x=1.
 
  • #12
wvcaudill2 said:
Ok, I see what you are saying now, but when I graphed the parabola, I also saw that coming from the right-hand side, the values of x were also positive up until x=1.

The parabola is negative between -1 and 1: http://www.wolframalpha.com/input/?i=y%3Dx^2-1
 
  • #13
Im still not seeing the big picture here. Can you write out exactly what this problem should look like with all the steps at least partially worked out?
 

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