What is the limit of |x+1| as x approaches -1?

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Homework Help Overview

The discussion revolves around evaluating the limit of the absolute value function |x+1| as x approaches -1. Participants are exploring the concept of absolute value and its implications for limits in calculus.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the definition of absolute value and its application in the limit problem. There are questions about the interpretation of |x+1| and whether the limit approaches a specific value. Some suggest breaking down the limit into left- and right-handed limits.

Discussion Status

The discussion is ongoing, with participants providing insights into the definition of absolute value and its relevance to the limit. There is a mix of interpretations and attempts to clarify the concept, but no consensus has been reached on the limit's value.

Contextual Notes

Some participants express confusion regarding the definition of absolute value and its application in the context of limits, indicating a need for further exploration of these foundational concepts.

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Homework Statement


Evaluate.
lim |x+1|
x-> -1


Homework Equations





The Attempt at a Solution


Not too sure what |x+1| means. I think it has something to do with an absolute value... would the answer be 2 then?
 
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What's the definition of absolute value? The answer is def. not 2
 
|x+1| is the absolute value of x+1. The easiest way to solve the problem is to realize that |x+1| = [(x+1)2]1/2.
 
NoMoreExams said:
What's the definition of absolute value? The answer is def. not 2

That's exactly my question. I still have no idea..
 
The absolute value of a number is the (positive) distance of the number from the origin, 0. In other words, if the number x is positive, the distance from the origin |x| is simply x - 0 = x. If x is a negative number, then the distance from the origin is 0 - x = -x, a positive number. In order to reconcile this with a single formula, we rigorously define the absolute value of a real number x to be |x| = \sqrt{x^2} which will always be positive, as the principal square root function is defined to be. Sometimes you will want to use this definition, other times it is easier to use the first definition:
|x| = \left\{\begin{array}{lr}x & ,x \geq 0\\ -x & ,x < 0\end{array}
The two definitions are equivalent.
Since you're considering the limit as x approaches -1 of |x+1|, which considers the behavior of the expression near |0|, you can either break it into left- and right-handed limits with the second definition, noting that the limit exists if and only if both left- and right-handed limits exist and are equivalent, or you can do the full limit with the square root definition, squaring the expression and studying the behavior of the principal square root as x approaches -1.
 
THe limit appraoches to 0 from both sides. derivative wouldn't exist at 0 since it's a corner.
 
Hi Lchan. Welcome to Physics Forums!
Lchan1 said:
derivative wouldn't exist at 0 since it's a corner.

Be careful with using the behavior of graphs in place of proof (they can be used to motivate a proof however). Ie., how do you define "corner" rigorously, so that given a function at a point, we can use the definition to decide whether or not there is a corner at that point?
 

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