1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits at absolute values?

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Evaluate.
    lim |x+1|
    x-> -1


    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Feb 6, 2009 #2
    What's the definition of absolute value? The answer is def. not 2
     
  4. Feb 7, 2009 #3
    |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.
     
  5. Feb 7, 2009 #4
    That's exactly my question. I still have no idea..
     
  6. Feb 7, 2009 #5
    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 [itex]|x| = \sqrt{x^2}[/itex] 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:
    [tex]|x| = \left\{\begin{array}{lr}x & ,x \geq 0\\ -x & ,x < 0\end{array}[/tex]
    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.
     
  7. Feb 7, 2009 #6
    THe limit appraoches to 0 from both sides. derivative wouldnt exist at 0 since it's a corner.
     
  8. Feb 7, 2009 #7
    Hi Lchan. Welcome to Physics Forums!
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limits at absolute values?
  1. Limit of Absolute Value (Replies: 11)

Loading...